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Maths Solve the Following Question.(5 Marks)

Statistics · Maharashtra Board STD 11 (MSBSHSE - Semi English Medium). 29 questions.

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Solve the Following Question.(5 Marks)

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29 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks
Find the coefficient of variation for the following data:
Size (in cms):
$10-15$
$15-20$
$20-25$
 $25-30$ $30-35$ $35-40$
No. of items:
 $2$
 
$8$
$20$
 $35$ $20$ $15$
Answer
$CI$ $f$ $x$ $\text{u}=\frac{\text{x}-\text{A}}{\text{h}}$ $fu$ $u^2$ $fu^2$
$10-15$ $2$ $12.5$ $-2$ $-4$ $4$ $8$
$15-20$ $8$ $17.5$ $-1$ $-8$ $1$ $8$
$20-25$ $20$ $22.5$ $0$ $0$ $0$ $0$
$25-30$ $35$ $27.5$ $1$ $35$ $1$ $35$
$30-35$ $20$ $32.5$ $2$ $40$ $4$ $80$
$35-40$ $15$ $37.5$ $3$ $45$ $9$ $135$
  $100$     $108$   $266$
Here, $N = 100, A = 22.5,  \sum\text{f}_\text{i}\text{u}_\text{i}=108,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=266$ and $h = 5$
$\therefore\text{Mean}=\overline{\text{x}}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\Rightarrow\overline{\text{x}}=22.5+5\Big(\frac{108}{100}\Big)=27.90$
$\text{Var}(\text{X})=\text{h}^2\bigg[\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg]$
$\text{Var}(\text{X})=25\bigg[\frac{266}{100}-\Big(\frac{108}{100}\Big)^2\bigg]=37.34$
$\therefore\text{S.D.}=\sqrt{\text{Var}(\text{X})}=\sqrt{37.34}=6.11$
Coefficient of variation $=\frac{\text{S.D.}}{\overline{\text{x}}}\times100=\frac{6.11}{27.90}\times100=21.9$
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Question 25 Marks
The mean and standard deviation of 20 observation are found to be 10 and 2 respectively. On reacheking it was found that an observation 8 was incorrect. Calculate the correct and standard deviation in each of the following cases:
  1. If wrong item is omitted.
  2. If it is replaced by 12.
Answer
$\text{n}=20,\overline{\text{X}}=10,\sigma=2$
$\therefore\overline{\text{x}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}$
$\Rightarrow\sum\text{x}_\text{i}=\text{n}\overline{\text{x}}=20\times10=200$
$\Rightarrow\text{Incorrected}\sum\text{x}_\text{i}=200$
and,
$\sigma=2$
$\Rightarrow\sigma^2=4$
$\Rightarrow\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\text{Mean})^2=4$
$\Rightarrow\frac{1}{20}\sum{\text{x}_\text{i}}^2-100=4$
$\Rightarrow\sum{\text{x}_\text{i}}^2=104\times20$
$\Rightarrow\sum{\text{x}_\text{i}}^2=2080.$
  1. When 8 is omitted from the data:
If 8 is omitted from the data, then 19 observation are left.
Now, Incorrected $\sum\text{x}_\text{i}=200$
⇒ Corrected $\sum{\text{x}_\text{i}}^2+8^2=2080$
⇒ Corrected $\sum{\text{x}_\text{i}}^2=2080-64$
⇒ Corrected $\sum\text{x}_\text{i}^2=2016$
$\therefore\text{Corrected}\ \text{mean}=\frac{192}{19}=10.10$
⇒ Corrected variance $=\frac{1}{19}\big(\text{corrected}\sum{\text{x}_\text{i}}^2\big)-\big(\text{Corrected}\ \text{mean}\big)^2$
⇒ Corrected variance $=\frac{2016}{19}-\Big(\frac{192}{19}\Big)^2$
Corrected variance $=\frac{38304-36864}{361}=\frac{1440}{361}$
$\therefore$ Corrected standard deviation $=\sqrt{\frac{1440}{361}}=\frac{12\sqrt{10}}{19}=1.997$
  1. When the incorrect observation 8 is replaced by 12:
We have, Incorrected $\sum\text{x}_\text{i}=200$
$\therefore\text{Corrected}\sum\text{x}_\text{i}=208-8+12=204$
and,
$\Rightarrow\text{Incorrected}\sum{\text{x}_\text{i}}^2=2080$
$\therefore\text{Corrected}\sum{\text{x}_\text{i}}^2=2080-8^2+12^2=2160$
Now, $\therefore\text{Corrected}\ \text{mean}=\frac{204}{20}=10.2$
Corrected variance $=\frac{1}{20}(\text{Corrected}\sum{\text{x}_\text{i}}^2)-(\text{Corrected}\ \text{mean})^2$
⇒ Corrected variance $=\frac{2160}{20}-\Big(\frac{204}{20}\Big)^2$
⇒ Corrected variance $=\frac{2160\times20-(204)^2}{(20)^2} $
⇒ Corrected variance $=\frac{43200-41616}{400}=\frac{1584}{400}$
$\therefore$ Corrected standard deviation $=\sqrt{\frac{1584}{400}}=\frac{\sqrt{396}}{10}=1.9899$
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Question 35 Marks
Find the standard deviation for the following data:
$x$
 $2$
$3$
$4$
$5$
$6$
 $7$
$f$
$4$
$9$
$16$
$14$
$11$
 $6$
Answer
$x_i$
$f_i$
$f_ix_i^2$
$2$
$4$
$16$
$3$
$9$
$81$
$4$
$16$
$256$
$5$
$14$
$350$
$6$
$11$
$396$
$7$
$6$
$294$
  $N = 60$ Total $= 1393$
Mean $=\frac{8+27+64+70+66+42}{60}=\frac{277}{60}=4.62$
Var $=\frac{1393}{60}-(4.62)^2=1.88$
SD $=\sqrt{1.88}=1.37$
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Question 45 Marks
Find the numberof observation lying between $\overline{\text{X}}-\text{M.D. }$and $\overline{\text{X}}-\text{M.D. }$ is the mean deviation from the mean.
$34, 66, 30, 38, 44, 50, 40, 60, 42, 51$
Answer
Let $\overline{\text{x}}$ be the mean of the data set.
$\overline{\text{x}}=\frac{34+66+30+38+44+50+40+60+42+51}{10}=45.5$
$\text{MD}=\frac{1}{\text{n}}\sum_\limits{\text{i}=1}^\text{n}|\text{d}_\text{i}|,\text{where}|\text{d}_\text{i}|=|\text{x}_\text{i}-\overline{\text{x}}|$
$x_i$
$|d_i| = |x_i - 45.5|$
$34$
$11.5$
$66$
$20.5$
$30$
$15.5$
$38$
$7.5$
$44$
$1.5$
$50$
$4.5$
$40$
$5.5$
$60$
$14.5$
$42$
$3.5$
$51$
$5.5$
Total
$90$
$\text{MD}=\frac{1}{10}\times90=9$
$\overline{\text{x}} - M.D. = 45.5 - 9 = 36.5$
Also, $\overline{\text{x}}+ M.D. = 45.5 + 9 = 54.5$
Hence, there are 6 observations between 36.5 and 54.5.
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Question 55 Marks
While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
Answer
Mean = 45
Variance = 16
n = 10
$\sum\text{x}_{\text{i}}=450$
Corrected Sum = 450 - 52 + 25 = 423
Corrected Mean = 42.3
Variance = 16
$16=\frac{\sum\text{x}_{\text{i}}^2}{10}-(45)^2$
Incorrected $\sum\text{x}_{\text{i}}^2=20410$
Corrected $\sum\text{x}_{\text{i}}^2=$
Incorrect $\sum\text{x}_{\text{i}}^2$ - (Sum of squares of incorrect value) + (Sum of squares of corrected value)
Corrected $\sum\text{x}_{\text{i}}^2=20410-2704+625=18331$
Corrected $\sigma=\sqrt{\frac{\text{Corrected}\sum\text{x}_\text{i}^2}{\text{n}}-(\text{Corrected Mean})^2}$
Corrected $\sigma=\sqrt{\frac{18331}{10}-(42.3)^2}=6.62$
Corrected Variance = 6.62 × 6.62 = 43.82
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Question 65 Marks
The mean and standard deviation of $6$ observation are $8$ and $4$ respectively. If each observation is multiplied by $3,$ find the new mean and new standard deviation of the resulting observation.
Answer
Mean = $\overline{\text{X}}=8$
$n = 6$
$\sigma=\text{S.D}=4$
If $x_1, x_2, ......x_6 $ are the given observation
$\overline{\text{X}}=\frac{1}{\text{n}}\times\sum_\limits{\text{i}=1}^6\text{x}_\text{i}$
$\Rightarrow8=\frac{1}{6}\times\sum_\limits{\text{i}=1}^6\text{x}_\text{i}$
Let $u_1, u_2.....u_6 $ be the new observation
$\Rightarrow u_i = 3x_i (for i = 1, 2, 3...6)$
$⇒$ Mean of new observation $=\overline{\text{U}}=\frac{1}{\text{n}}\times\sum_\limits{\text{i}=1}^6\text{u}_\text{i}$
$=\frac{1}{6}\times\sum_\limits{\text{i}=1}^63\text{x}_\text{i}$
$=3\times\frac{1}{6}\times\sum_\limits{\text{i}=1}^6\text{x}_\text{i}$
$=3\ \overline{\text{X}}$
$=3\times8$
$=24$
$\text{SD}=\sigma_\text{x}=4$
${\sigma_\text{x}}^2$ = Variance $X$
$\therefore$ Variance $X = 16$
$\Rightarrow\frac{1}{6}\sum_\limits{\text{i}=1}^6(\text{x}_\text{i}-\overline{\text{X}})^2=16.....(1)$
$\text{Variance}\ (\text{U})={\sigma_\text{u}}^2=\frac{1}{6}\sum_\limits{\text{i}=1}^6(\text{u}_\text{i}-\overline{\text{U}})^2$
$=\frac{1}{6}\times\sum_\limits{\text{i}=1}^6(3\text{x}_\text{i}-3\overline{\text{X}})^2$
$=3^2\times\frac{1}{6}\sum_\limits{\text{i}=1}^6(\text{x}_\text{i}-\overline{\text{X}})^2$
$=9\times16$
$\sigma_\text{u}=\sqrt{\text{Variance}\ (\text{U})}$
$=\sqrt{9\times16}$
$=12$
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Question 75 Marks
Find the numberof observation lying between $\overline{\text{X}}-\text{M.D. }$and $\overline{\text{X}}+\text{M.D. }$ is the mean deviation from the mean.
$22, 24, 30, 27, 29, 31, 25, 28, 41, 42$
Answer
Let $\overline{\text{x}}$ be the mean of the data set.
$\overline{\text{x}}=\frac{22+24+30+27+29+31+25+28+41+42}{10}=29.9=$
$x_i$
$|d_i| = |x_i - 29.9|$
$22$
$7.9$
$24$
$5.9$
$30$
$0.1$
$27$
$2.9$
$29$
$0.9$
$31$
$1.1$
$25$
$4.9$
$28$
$1.9$
$41$
$11.9$
$42$
$12.1$
Total
$48.8$
$\text{MD}=\frac{1}{10}\times48.8=4.88$
$\overline{\text{x}} - M.D. = 29.9 - 4.88 = 25.02,$
and, $\overline{\text{x}} + M.D. = 29.9 + 4.88 = 34.78$
There are $5$ observation between $25.02$ and $34.78.$
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Question 85 Marks
Show that the two formula for the standard deviation of ungrouped data
$\sigma=\sqrt{\frac{1}{\text{n}\sum(\text{x}_\text{i}-\overline{\text{X}})^2}}\text{and}\ \sigma^{'}=\sqrt{\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-\overline{\text{X}^2}}$ are equivalent, where $ \overline{\text{X}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}.$
Answer
$\sigma=\sqrt{\frac{1}{\text{n}}\sum(\text{x}_\text{i}-\overline{\text{X}})^2}$
$=\sqrt{\frac{1}{\text{n}}\sum\big({\text{x}_\text{i}^2}-2{\text{x}_\text{i}}\overline{\text{X}}+\overline{\text{X}}\big)^2}$
$=\sqrt{\frac{1}{\text{n}}\sum{\text{x}^2_\text{i}}-\frac{1}{\text{n}}\sum2\text{x}_\text{i}\overline{\text{X}}+\frac{1}{\text{n}}\sum\overline{{\text{X}}}^2}$
$=\sqrt{\frac{1}{\text{n}}\sum{\text{x}}^2_\text{i}-\frac{1}{\text{n}}\times2\overline{\text{X}}\sum\text{x}_\text{i}+\frac{1}{\text{n}}\times\overline{\text{X}}^2\sum1}$
$=\sqrt{\frac{1}{\text{n}}\sum{\text{x}^2_\text{i}}-\frac{1}{\text{n}}\times2\overline{\text{X}}\times\text{n}\overline{\text{X}}+\frac{1}{\text{n}}\times\overline{\text{X}}^2\times\text{n}}$
$=\sqrt{\frac{1}{\text{n}}\sum\text{x}^2_\text{i}-2\overline{\text{X}}^2+\overline{\text{X}}^2}$
$=\sqrt{\frac{1}{\text{n}}\sum\text{x}^2_\text{i}-\overline{\text{X}}^2}$
$=\sigma^{'}$
Hence, the formula $\sigma=\sqrt{\frac{1}{\text{n}}\sum(\text{x}_\text{i}-\overline{\text{X}})^2}\ \text{and}\ \sigma^{'}=\sqrt{\frac{1}{\text{n}}\sum\text{x}^2_\text{i}-\overline{\text{X}}^2}$ are equivalent, where $\overline{\text{X}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}.$
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Question 95 Marks
The mean and variance of $8$ observation are $9$ and $9.25$ respectively. If six of the observation are $6, 7, 10, 12, 12$ and $13,$ find the remaining two observation.
Answer
Let $x$ and $y$ be the remaining two observations. Then,
Mean $= 9$
$\Rightarrow\frac{6+7+10+12+12+13+\text{x}+\text{y}}{8}=9$
$\Rightarrow 60 + x + y =72$
$\Rightarrow x + y = 12.........(i)$
Variance $= 9.25$
$\Rightarrow\frac{1}{8}\big(6^2+7^2+10^2+12^2+12^2+13^2+\text{x}^2+\text{y}^2\big)-(\text{Mean})^2=9.25$
$\Rightarrow\frac{1}{8}\big(36+49+100+144+144+169+\text{x}^2+\text{y}^2\big)-81=9.25$
$\Rightarrow 642 + x^2 + y^2 = 722$
$\Rightarrow x^2 + y^2 = 80........(ii)$
Now, $(x + y)^2 + (x - y)^2 = 2(x^2 + y^2)$
$\Rightarrow 144 + (x - y)^2= 2 \times 80$
$\Rightarrow x - y = 16$
$\Rightarrow\text{x}-\text{y}=\pm4$
if $x - y = 4,$ then $x + y =12$ and $x - y = 4 \Rightarrow x = 8, y = 4$
if $x - y = -4,$ then $x + y =12$ and $x - y = -4 \Rightarrow x = 4, y = 8$
Hence, the remaining two observation are $4$ and $8.$
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Question 105 Marks
The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observation were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observation were omitted.
Answer
We have, $\text{n}=100,\overline{\text{x}}=20\ \text{and}\ \sigma=3$
Since $\overline{\text{x}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}$
$\Rightarrow\sum\text{x}_\text{i}=\text{n}\overline{\text{x}}=20\times100=2000$
$\Rightarrow\text{Incorrect}\ \sum\text{x}_\text{i}=2000$
and,
$\sigma=3$
$\Rightarrow\sigma^2=9$
$\Rightarrow\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\text{Mean})^2=9$
$\Rightarrow\frac{1}{100}\sum{\text{x}_\text{i}}^2-400=9$
$\Rightarrow\sum{\text{x}_\text{i}}^2=409\times100$
$\Rightarrow\text{Incorrect}\sum{\text{x}_\text{i}}^2=40900.$
When the incorrect observation 21, 21, 18 are omitted from the data:
n = 97
Now, $\text{Incorrect}\sum\text{x}_\text{i}=2000$
$\Rightarrow\sum{\text{x}_\text{i}}=2000-21-21-18=1940$
and,
$\text{Incorrect}\sum{\text{x}_\text{i}}^2=40900$
$\Rightarrow\text{Corrected}\sum{\text{x}_\text{i}^2}=40900-21^2-21^2-18^2$
$\Rightarrow\text{Corrected}\sum{\text{x}_\text{i}^2}=40900-1206$
$\Rightarrow\text{Corrected}\sum{\text{x}_\text{i}^2}=39694$
$\therefore\text{Corrected}\ \text{mean}=\frac{1940}{97}=20$
$\Rightarrow\text{Corrected}\ \text{variavce}=\frac{1}{97}\big(\text{Corrected}\sum{\text{x}_\text{i}}^2\big)-\big(\text{Corrected}\ \text{mean}\big)^2$
$\Rightarrow\text{Corrected}\ \text{variavce}=\frac{39694}{97}-(20)^2=409.22-400=9.22$
$​​\therefore\text{Corrected}\ \text{standard}\ \text{deviation}=\sqrt{9.22}=3.04$
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Question 115 Marks
Calculate the mean, variance and standard deviation of the following frequency distribution.
Class: $1-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency: $11$ $29$ $18$ $4$ $5$ $3$
Answer
Class interval
$f_i$
Mid-value $x_i$
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-35}{10}$
$f_iu_i$
$u_i^2$
$f_iu_i^2$
$0-10$
$11$
$5$
$-3$
$-33$
$9$
$99$
$10-20$
$29$
$15$
$-2$
$-58$
$4$
$116$
$20-30$
$18$
$25$
$-1$
$-18$
$1$
$18$
$30-40$
$4$
$35$
$0$
$0$
$0$
$0$
$40-50$
$5$
$45$
$1$
$5$
$1$
$5$
$50-60$
$3$
$55$
$2$
$6$
$4$
$12$
 
$\text{N}=\sum\text{f}_\text{i}=70$
 
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=-98$
 
$\sum\text{f}_\text{i}\text{u}_\text{i}^2=250$
$N = 70, \sum\text{f}_\text{i}\text{u}_\text{i}^2=250, A = 35$ and $h = 10$
$\text{Mean}=\text{A}+\text{h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\text{Mean}=35+10\Big(\frac{-98}{70}\Big)=-21$
$\text{Var}\big(\text{X}\big)=\text{h}^2\bigg\{\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg\}$
$\text{Var}\big(\text{X}\big)=100\bigg\{\Big(\frac{1}{70}\times250\Big)-\Big(\frac{1}{70}\times(-98)\Big)^2\bigg\}$
$\text{Var}\big(\text{X}\big)=100\big\{3.57-1.96\big\}=161$
$\text{SD}=\sqrt{\text{Var}(\text{X})}=\sqrt{161}=12.7$
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Question 125 Marks
calculate the mean deviation from the mean for the following data:
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answer
$\text{Mean}=\frac{1}{\text{n}}\sum|\text{x}_\text{i}|=\frac{168}{12}=14$
Calculation of Mean Deviation
X-values
Deviation From Mean
13
1
17
3
16
2
14
0
11
3
13
1
10
4
16
2
11
3
18
4
12
2
17
3
Total
28
We have,
$\sum|\text{x}_\text{i}-14|=\sum\text{d}_\text{i}=28$
$\therefore\text{M.D}=\frac{1}{\text{n}}\sum|\text{d}_\text{i}|=\frac{1}{12}[28]=2.33$
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Question 135 Marks
Table below shows the frequency $f$ with which $'x'$ alpha particles radiated from a diskette:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$
$f$ $51$ $203$ $383$ $525$ $532$ $408$ $273$ $139$ $43$ $27$ $10$ $4$ $2$
Calculate the mean and variance.
Answer
Mean, $\overline{\text{x}}=\frac{\sum\text{f}_{\text{i}}\text{x}_{\text{i}}}{\sum\text{f}_{\text{i}}}=\frac{10078}{2600}=3.88$
$x_i$
$f_i$
$f_ix_i$
$\text{x}_{\text{i}}-\overline{\text{X}}$
$\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2$
$\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2$
$0$
$51$
$0$
$-3.88$
$15.05$
$767.55$
$1$
$203$
$203$
$-2.88$
$8.29$
$1682.87$
$2$
$383$
$766$
$-1.88$
$3.53$
$1351.99$
$3$
$525$
$1575$
$-0.88$
$0.77$
$404.25$
$4$
$532$
$2128$
$0.12$ 
$0.014$
$7.448$
$5$
$408$
$2040$
$1.12$
$1.25$
$510$
$6$
$273$
$1638$
$2.12$
$4.49$
$1225.77$
$7$
$139$
$973$
$3.12$
$9.73$
$1352.47$
$8$
$43$
$344$
$4.12$
$16.97$
$729.71$
$9$
$27$
$243$
$5.12$
$26.21$
$707.67$
$10$
$10$
$100$
$6.12$
$37.45$
$374.5$
$11$
$4$
$44$
$7.12$
$50.69$
$202.76$
$12$
$2$
$24$
$8.12$
$65.93$
$131.86$
 
$\sum\text{f}_{\text{i}}=\text{N}=2600$
$\sum\text{f}_{\text{i}}\text{x}_{\text{i}}=10078$
 
 
$\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2=9448.848$
Variance, $\sigma^2=\frac{\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{X}}\big)^2​​}{{\text{N}}}=\frac{9448.848}{2600}=3.63$
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Question 145 Marks
Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Answer
Mean = 40
SD = 10
n = 100
$\sum\text{x}_{\text{i}}=40\times100=4000$
Corrected Sum = 4000 - 30 - 70 + 3 + 27 = 3930
Corrected Mean = $\frac{3930}{100}=39.3$
Variance = 100
$100=\frac{\sum\text{x}_{\text{i}}^2}{100}-(40)^2$
Incorrected $\sum\text{x}_{\text{i}}^2=170000$
Corrected $\sum\text{x}_{\text{i}}^2=$
Incorrect $\sum\text{x}_{\text{i}}^2$ - (Sum of squares of incorrect value) + (Sum of squares of corrected value)
Corrected $\sum\text{x}_{\text{i}}^2=170000-(900+4900)+(9+729)$
Corrected $\sum\text{x}_{\text{i}}^2=164938$
Corrected $\sigma=\sqrt{\frac{\text{Corrected}\sum\text{x}_\text{i}^2}{\text{n}}-(\text{Corrected Mean})^2}$
Corrected $\sigma=\sqrt{\frac{164938}{100}-(39.3)^2}=10.24$
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Question 155 Marks
Find the mean, and standard deviation for the following data:
Marks:
$2$
$3$
$4$
$5$
$6$
$7$
$8$
$9$
$10$
$11$
$12$
$13$
$14$
$15$
$16$
Frequency:
$1$
$6$
$6$
$8$
$8$
$2$
$2$
$3$
$0$
$2$
$1$
$0$
$0$
$0$
$1$
Answer
$x_i$ $f_i$ $f_ix_i$ $f_ix_i^2$
$2$ $1$ $2$ $4$
$3$ $6$ $18$ $54$
$4$ $6$ $24$ $96$
$5$ $8$ $40$ $200$
$6$ $8$ $48$ $288$
$7$ $2$ $14$ $98$
$8$ $2$ $16$ $128$
$9$ $3$ $27$ $243$
$10$ $0$ $0$ $0$
$11$ $2$ $22$ $242$
$12$ $1$ $12$ $144$
$13$ $0$ $0$ $0$
$14$ $0$ $0$ $0$
$15$ $0$ $0$ $0$
$16$ $1$ $16$ $256$
  $N = 40$ Total $= 239$ Total $= 1753$
Mean $=\frac{239}{40}=5.975$
Var $=\frac{1753}{40}-(5.975)^2=8.12$
SD $=\sqrt{8.12}=2.85$
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Question 165 Marks
Following are the marks obtained, out of $100,$ by two students Ravi and Hashina in $10$ tests:
Ravi: $25$ $50$ $45$ $30$ $70$ $42$ $36$ $48$ $35$ $60$
Hashina: $10$ $70$ $50$ $20$ $95$ $55$ $42$ $60$ $48$ $80$
Who is more intelligent and who is more consistent?
Answer
For Ravi:
Marks $(x_i)$
$d_i = x_i - 45$
$d_i^2$
$25$
$-20$
$400$
$50$
$5$
$25$
$45$
$0$
$0$
$30$
$-15$
$225$
$70$
$25$
$625$
$42$
$-3$
$9$
$36$
$-9$
$81$
$48$
$3$
$9$
$35$
$-10$
$100$
$60$
$15$
$225$
 
$\sum\text{d}_\text{i}=-9$
$\sum\text{d}_\text{i}^2=1699$
Mean, $\overline{\text{X}}_\text{R}=\text{A}+\frac{\sum\text{d}_\text{i}}{10}=45+\frac{(-9)}{10}=44.1$
Standard deviation, $\sigma_\text{R}=\sqrt{\frac{\sum\text{d}_\text{i}^2}{10}-\Big(\frac{\sum\text{d}_\text{i}}{10}\Big)^2}=\sqrt{\frac{1699}{10}-\Big(\frac{-9}{10}\Big)^2}=\sqrt{169.09}=13.003$
Coefficient of variation $=\frac{\sigma_\text{R}}{\overline{\text{X}}_\text{R}}\times100=\frac{13.003}{44.1}\times100=29.49$
For Hashina:
Marks $(x_i)$
$d_i = x_i - 55$
$d_i^2$
$10$
$-45$
$2025$
$70$
$15$
$625$
$50$
$-5$
$25$
$20$
$-35$
$1225$
$95$
$40$
$1600$
$55$
$0$
$0$
$42$
$-13$
$169$
$60$
$5$
$25$
$48$
$-7$
$49$
$80$
$25$
$625$
 
$\sum\text{d}_\text{i}=-20$
$\sum\text{d}_\text{i}^2=6368$
Mean, $\overline{\text{X}}_\text{H}=\text{A}+\frac{\sum\text{d}_\text{i}}{10}=55+\frac{(-20)}{10}=53$
Standard deviation, $\sigma_\text{H}=\sqrt{\frac{\sum\text{d}_\text{i}^2}{10}-\Big(\frac{\sum\text{d}_\text{i}}{10}\Big)^2}=\sqrt{\frac{6368}{10}-\Big(\frac{-20}{10}\Big)^2}=\sqrt{632.8}=25.16$
Coefficient of variation $=\frac{\sigma_\text{H}}{\overline{\text{X}}_\text{H}}\times100=\frac{25.16}{53}\times100=47.47$
Since the coefficient of variation in mark obtained by Hashima is greater than the coefficient of variation in mark obtained by Ravi, so Hashina is more consistent and intelligent.
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Question 175 Marks
Find the mean, and standard deviation for the following data:
Year render: $10$ $20$ $30$ $40$ $50$ $60$
No. of persons(cumulative): $15$ $32$ $51$ $78$ $97$ $109$
Answer
$x$ Cum Freq $f_i$ $f_ix_i$ $f_ix_i^2$
$10$ $15$ $15$ $150$ $1500$
$20$ $32$ $17$ $340$ $6800$
$30$ $51$ $19$ $570$ $17100$
$40$ $78$ $27$ $1080$ $43200$
$50$ $97$ $19$ $950$ $47500$
$60$ $109$ $12$ $720$ $43200$
    $N = 109$ Total $= 3810$ Total $= 159300$
Mean $=\frac{3810}{109}=34.95$
Var $=\frac{159300}{109}-(34.95)^2=239.96$
SD $=\sqrt{239.96}=15.49$
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Question 185 Marks
From the prices of shares $X$ and $Y$ given below: find out which is more stable in value:
$x$ $35$ $54$ $52$ $53$ $56$ $58$ $52$ $50$ $51$ $49$
$y$ $108$ $107$ $105$ $105$ $106$ $107$ $104$ $103$ $104$ $101$
Answer
$x$ d$ = (x -$ Mean$)$ $d^2$
$35$ $-13$ $169$
$24$ $-24$ $576$
$52$ $4$ $16$
$53$ $5$ $25$
$56$ $8$ $64$
$58$ $10$ $100$
$52$ $4$ $16$
$50$ $2$ $4$
$51$ $3$ $9$
$49$ $1$ $1$
$480$   $980$
$\overline{\text{x}}=\frac{1}{\text{N}}\sum\text{x}_\text{i}=\frac{1}{10}[480]=48$
$\text{Var}(\text{X})=\frac{1}{\text{N}}\Big\{\sum\big(\text{x}_\text{i}-\overline{\text{x}}\big)^2\Big\}=\frac{1}{10}(980)=98$
$\therefore\text{S.D.}(\text{X})=\sqrt{\text{Var}(\text{X})}=\sqrt{98}=9.9$
Coefficient of variation $=\frac{\text{S.D.}}{\overline{\text{x}}_1}\times100=\frac{9.9}{48}\times100=20.6$
$x$ $d = (x -$ Mean$)$ $d^2$
$35$ $-13$ $169$
$24$ $-24$ $576$
$52$ $4$ $16$
$53$ $5$ $25$
$56$ $8$ $64$
$58$ $10$ $100$
$52$ $4$ $16$
$50$ $2$ $4$
$51$ $3$ $9$
$49$ $1$ $1$
$480$   $980$
$\overline{\text{x}}=\frac{1}{\text{N}}\sum\text{x}_\text{i}=\frac{1}{10}[1050]=105$
$\text{Var}(\text{X})=\frac{1}{\text{N}}\Big\{\sum\big(\text{x}_\text{i}-\overline{\text{x}}\big)^2\Big\}=\frac{1}{10}(40)=4$
$\therefore\text{S.D}(\text{X})=\sqrt{\text{Var}(\text{X})}=\sqrt{4}=2$
Coefficient of variation for shares $\text{Y}=\frac{\text{S.D.}}{\overline{\text{x}}_1}\times100=\frac{2}{105}\times100=1.90$
Since the coefficient of variation for share $Y$ is smaller than the coefficient of variation for share $X,$ they are more stable.
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Question 195 Marks
From the data given below state which group is more variable, $G_1$ or $G_2$?
Marks
$10-20$
$20-30$
$30-40$
 $40-50$ $50-60$ $60-70$ $70-80$
Group $G_1$
$9$
$17$
$32$
 $33$ $40$ $10$ $9$
Group $G_2$
$10$
$20$
$30$
 $25$ $43$ $15$ $7$
Answer
Let's first find the coefficient of variable for Group $G_1$
CI f x $\text{u}=\frac{\text{x}-\text{A}}{\text{h}}$ fu $u^2$ $fu^2$
$10-20$ $9$ $15$ $-3$ $-27$ $9$ $81$
$20-30$ $17$ $25$ $-2$ $-34$ $4$ $68$
$30-40$ $32$ $35$ $-1$ $-32$ $1$ $32$
$40-50$ $33$ $45$ $0$ $0$ $0$ $0$
$50-60$ $40$ $55$ $1$ $40$ $1$ $40$
$60-70$ $10$ $65$ $2$ $20$ $4$ $40$
$70-80$ $9$ $75$ $3$ $27$ $9$ $81$
  $150$     $-6$   $342$
Here, $N = 150, A = 45$, $\sum\text{f}_\text{i}\text{u}_\text{i}=-6,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=342$ and $h = 10$
$\therefore\text{Mean}=\overline{\text{x}}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\Rightarrow\overline{\text{x}}=45+10\Big(\frac{-6}{150}\Big)=44.6$
$\text{Var}(\text{X})=\text{h}^2\bigg[\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg]$
$\text{Var}(\text{X})=100\bigg[\frac{342}{150}-\Big(\frac{-6}{150}\Big)^2\bigg]=227.84$
$\therefore\text{S.D.}=\sqrt{\text{Var}(\text{X})}=\sqrt{227.84}=15.09$
Coefficient of variation $=\frac{\text{S.D.}}{\overline{\text{x}}}\times100=\frac{15.09}{44.6}\times100=33.83$
Now, Let's first find the coefficient of variable for Group $G_2$
CI f x $\text{u}=\frac{\text{x}-\text{A}}{\text{h}}$ $fu$ $u^2$ $fu^2$
$10-20$ $10$ $15$ $-3$ $-30$ $9$ $902$
$20-30$ $20$ $25$ $-2$ $-40$ $4$ $80$
$30-40$ $30$ $35$ $-1$ $-30$ $1$ $30$
$40-50$ $25$ $45$ $0$ $0$ $0$ $0$
$50-60$ $43$ $55$ $1$ $43$ $1$ $43$
$60-70$ $15$ $65$ $2$ $30$ $4$ $60$
$70-80$ $7$ $75$ $3$ $21$ $9$ $63$
  $150$     $-6$   $366$
Here, $N = 150, A = 45$, $\sum\text{f}_\text{i}\text{u}_\text{i}=-6,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=366$ and $h = 10$
$\therefore\text{Mean}=\overline{\text{x}}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)$
$\Rightarrow\overline{\text{x}}=45+10\Big(\frac{-6}{150}\Big)=44.6$
$\text{Var}(\text{X})=\text{h}^2\bigg[\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg]$
$\text{Var}(\text{X})=100\bigg[\frac{366}{150}-\Big(\frac{-6}{150}\Big)^2\bigg]=243.84$
$\therefore\text{S.D.}=\sqrt{\text{Var}(\text{X})}=\sqrt{227.84}=15.62$
Coefficient of variation $=\frac{\text{S.D.}}{\overline{\text{x}}}\times100=\frac{15.09}{44.6}\times100=35.02$
$\therefore$ Group $G_2$ is more variable.
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Question 205 Marks
The lengths (in cm) of 10 rods in a shop are given below:
$40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2$
  1. Find mean deviation from median
  2. Find mean deviation from the mean also.
Answer
First arrange the given numbers in assending order
write these number in assending order
$40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2$
we get $15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0$
Clearly, $\text{Median}=\frac{40.0+52.3}{2}=46.15$
Let $\overline{\text{x}}$ be the mean of given data , we get
$\overline{\text{x}}=\frac{15.2+27.9+ 30.2+32.5+40.0+52.3+52.8+55.2+72.9+79.0}{10}=45.8$
Calculation of mean Deviations from mean and median
$x_i$ $|d_i| = |x_i - 46.15|$ $|d_i| = |x_i - 45.8|$
$40.0$ $6.15$ $5.8$
$52.3$ $6.15$ $6.5$
$55.2$ $9.05$ $9.4$
$72.9$ $26.75$ $27.1$
$52.8$ $6.65$ $7$
$79.0$ $32.85$ $33.2$
$32.5$ $13.65$ $13.3$
$15.2$ $30.95$ $30.6$
$27.9$ $19.25$ $17.9$
$30.2$ $15.95$ $15.6$
Total $167.4$ $166.4$
  1. $\text{M.D}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{167.4}{10}=16.74$
  2. $\text{M.D}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{166.4}{10}=16.64$
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Question 215 Marks
Life of bulbs produced by two factories $A$ and $B$ are given below:
Length of life $($in hours$):$ $550-650$ $650-750$ $750-850$ $850-950$ $950-1050$
Factory $A: ($Number of bulbs$)$ $10$ $22$ $52$ $20$ $16$
Factory $B: ($Number of bulbs$)$ $8$ $60$ $24$ $16$ $12$
The bulbs of which factory are more consistent from the point of view of length of life?
Answer
Factor A:
Length of life
Mid value $x_i$
$f_i$
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-800}{100}$
$f_iu_i$
$f_iu_i^2$
$550-650$
$600$
$10$
$-2$
$-20$
$40$
$650-750$
$700$
$22$
$-1$
$-22$
$22$
$750-850$
$800$
$52$
$0$
$0$
$0$
$850-950$
$900$
$20$
$1$
$20$
$20$
$950-1050$
$1000$
$16$
2
$32$
$64$
 
 
$\text{N}=\sum\text{f}_\text{i}=120$
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=10$
$\sum\text{f}_\text{i}\text{u}_\text{i}^2=146$
 
 
N = 120, $\sum\text{f}_\text{i}\text{u}_\text{i}=10,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=146, A = 800$ and $h = 100$
$\overline{\text{x}}_\text{A}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)=800+100\Big(\frac{10}{120}\Big)=808.33$
$\sigma_\text{A}^2=\text{h}^2\bigg\{\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg\}$
$\sigma_\text{A}^2=10000\bigg\{\Big(\frac{1}{120}\times146\Big)-\Big(\frac{1}{120}\times(10)\Big)^2\bigg\}$
$\sigma_\text{A}^2=10000(1.2166-0.0069)=12097$
$\Rightarrow\sigma_\text{A}^2=\sqrt{12097}=109.98\approx110$
Factor B:
Length of life
Mid value $x_i$
$f_i$
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-800}{100}$
$f_iu_i$
$f_iu_i^2$
$550-650$
$600$
$8$
$-2$
$-16$
$32$
$650-750$
$700$
$60$
$-1$
$-60$
$60$
$750-850$
$800$
$24$
$0$
$0$
$0$
$850-950$
$900$
$16$
$1$ 
$16$
$16$
$950-1050$
$1000$
$12$
$2$
$12$
$48$
 
 
$\text{N}=\sum\text{f}_\text{i}=120$
 
$\sum\text{f}_\text{i}\text{u}_\text{i}=-48$
$\sum\text{f}_\text{i}\text{u}_\text{i}^2=156$
 
$N = 120, \sum\text{f}_\text{i}\text{u}_\text{i}=-48,\ \sum\text{f}_\text{i}\text{u}_\text{i}^2=156,$ $A = 800$ and $h = 100$
$\overline{\text{x}}_\text{B}=\text{A+h}\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)=800+100\Big(\frac{-48}{120}\Big)=760$
$\sigma_\text{B}^2=\text{h}^2\bigg\{\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}^2\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_\text{i}\text{u}_\text{i}\Big)^2\bigg\}$
$\sigma_\text{A}^2=10000\bigg\{\Big(\frac{1}{120}\times156\Big)-\Big(\frac{1}{120}\times(-48)\Big)^2\bigg\}$
$\sigma_\text{B}^2=10000(1.3-0.16)=11400$
$\Rightarrow\sigma_\text{B}=\sqrt{11400}=106.77\approx107$
Bulbs of factory A are more consistent from the point of view of life.
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Question 225 Marks
Find the standard deviation for the following distribution:
x 4.5 14.5 24.5 34.5 44.5 54.5 64.5
f 1 5 12 22 17 9 4
Answer
x f fx x-mean $(x-mean)^2$ $f(x-mean)^2$
4.5 1 4.5 -33.14 1098.45 1098.45
14.5 5 72.5 -23.14 535.59 2677.96
24.5 12 294 -13.14 172.73 2072.82
34.5 22 759 -3.14 9.88 217.31
44.5 17 756.5 6.86 47.02 799.35
54.5 9 490.5 16.86 284.16 2557.47
64.5 4 258 26.86 721.31 2885.22
  N = 70 2635     12308.57
Here, N = 70, $\sum\text{f}_{\text{i}}\text{x}_{\text{i}}=2635$
$\therefore\overline{\text{x}}=\frac{1}{\text{N}}\big(\sum\text{f}_{\text{i}}\text{x}_{\text{i}}\big)=\frac{2635}{70}=37.64$
we have, $\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{x}}\big)^2=12308.57$
$\therefore\text{ver}(\text{x})=\frac{1}{\text{N}}\Big[\sum\text{f}_{\text{i}}\big(\text{x}_{\text{i}}-\overline{\text{x}}\big)^2\Big]\\=\frac{12308.57}{70}=175.84$
$\text{S.D.}=\sqrt{\text{ver}(\text{x})}=\sqrt{175.84}=13.26$
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Question 235 Marks
The mean of 5 observation is 4.4 and their variance is 8.24. If three of the observation are 1, 2 and 6, find the other two observation.
Answer
Let the other two be x and y
1 + 2 + 6+ x + y = 5* 4.4 because of the mean
x + y = 13
Variance $=\frac{[(1-4.4)^2+(2-4.4)^2+(6-4.4)^2+(\text{x}-4.4)^2+(\text{y}-4.4)^2]}{5}$
Hence
11.56 + 5.76 + 2.56 + (x - 4.4)^2 + (y4.4)^2 = 41.2
(x - 4.4)^2 + (y - 4.4)^2 = 21.32
Solve simuitaneously
(x - 4.4)^2 + (13 - x - 4.4)^2 = 21.32
(x- 4.4)^2 + (8.6 - x)^2 = 21.32
x^2 - 8.8x + 19.36 + 73.96 - 17.2x + x^2 = 21.32
2x^2 - 26x + 72 = 0
x^2 - 13x + 36 = 0
(x - 4)(x - 9) = 0
x = 4 or x = 9
If x = 4, y = 9 and
The other two observation are4 and 9.
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Question 245 Marks
Find the mean and variance of frequency distribution given below:
$x_i$ $1\leq\text{x}<3$ $3\leq\text{x}<5$ $5\leq\text{x}<7$ $7\leq\text{x}<10$
$f_1$ 6 4 5 1
Answer
$x_i$ Midpoint value $\left(y_i\right)$ $y_i^2$ $f_i$ $f_i y_i$ $f_i y_i^2$
1-3
2
4
6
12
24
3-5
4
16
4
16
64
5-7
6
36
5
30
180
7-10
8.5
72.25
1
8.5
72.25
 
 
 
$\text{N}=\sum\text{f}_\text{i}=16$
$\sum\text{f}_\text{i}\text{y}_\text{i}=66.5$
$\sum\text{f}_\text{i}\text{y}_\text{i}^2=340.25$
Therfore,
$\text{Mean}=\frac{\sum\text{f}_{\text{i}}\text{y}_{\text{i}}}{\sum\text{f}_{\text{i}}}=\frac{66.5}{16}=4.16$
$\text{Variance}=\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{y}_{\text{i}}^2\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{y}_{\text{i}}\Big)^2$
$\text{Variance}=\frac{1}{16}\times340.25-\Big(\frac{1}{16}\times66.5\Big)^2$
$\text{Variance}=21.26-17.22=4.04$
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Question 255 Marks
The mean and standard deviation of 100 observation were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?
Answer
 We have,
$n =100, \overline{X}=40, \sigma=5.1$
$\therefore \overline{X}=\frac{1}{n} \sum x_{i}=\overline{X}=100 \times 40=4000$ Corrected $\sum x _{ i }=$ Incorrected $\sum x _{ i }-$ (sum of incorrect values) + (sum of correct values) $=4000-50+40=3990$
$\therefore$ Corrected mean $=\frac{\text { corrected } \sum x _1}{ n }=\frac{3990}{100}=39.9$ Now $\sigma=5.1$
$\Rightarrow 5.1^2=\frac{1}{100}\left(\sum x_{i}^2\right)-\left(\frac{1}{100} \sum x_{i}\right)^2$
$\Rightarrow 26.01=\frac{1}{100}\left(\sum x_{i}^2\right)-\left(\frac{4000}{100}\right)^2$
$\Rightarrow 26.01=\frac{1}{100}\left(\sum x_{i}^2\right)-1600$
$\sum x_{i}^2=100 \times 1626.01=162601$
$\text { Incorrect } \sum x_{i}^2=162601$
$\text { corrected } \sum x_{i}^2=\left(incorrected \sum x_{i}^2\right)-(\text { sum of squers of incorrect values })+(\text { sum of squers of correct values) }$
$=162601-(50)^2+(40)^2=161701$
$\text { so, Corrected } \sigma=\sqrt{\frac{1}{n} \sum x_{i}^2-\left(\frac{1}{n} \sum x_{i}\right)^2}=\sqrt{\frac{161701}{100}-\left(\frac{3990}{100}\right)^2}$
$=\sqrt{1617.01-1592.01}=5$
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Question 265 Marks
The weight of coffee in 70 jars is shown in the following table:
Weight (in grams): 200-201 201-202 202-203 203-204 204-205 205-206
Frequency: 13 27 18 10 1 1
Determine the variance and standard deviation of the above distribution.
Answer
Weight (in grams)
 Mid-values $\left( x _{ i }\right)$ Frequency $\left( f _{ i }\right)$ $d _{ i }= x _{ i }-202.5$ $d_{ i }^2$ $f _{ i } d _{ i }$ $f _{ i } d _{ i }^2$
200-201
 
200.5
  
13
-2
4
-26
52
201-202
201.5
27
-1
1
-27
27
202-203
202.5
18
0
0
0
0
203-204
203.5
10
1
1
10
10
204-205
204.5
1
2
 
4
2
4
 
205-206
205.5
1
3
9
3
9
 
 
$\text{N}=\sum\text{f}_\text{i}=70$
 
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=-38$
$\sum\text{f}_\text{i}\text{d}_\text{i}^2=102$
Now,
Variance, $\sigma^2$
$=\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{d}_{\text{i}}^2\Big)-\Big(\frac{1}{\text{N}}\sum\text{f}_{\text{i}}\text{d}_{\text{i}}\Big)^2$
$=\Big(\frac{1}{70}\times102\Big)-\Big(\frac{1}{70}\times(-38)\Big)^2$
$=1.457-0.295$
$=1.162\text{gm}$
Standard deviation, $\sigma=\sqrt{\text{Variance}}=\sqrt{1.162}=1.08\text{gm}$
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Question 275 Marks
Find the standard deviation for the following data:
x
3
8
13
18
23
f
7
10
15
10
6
Answer
x f fx x-mean $(x-mean)^2$ $f(x-mean)^2$
3 7 21 -9.79 95.88 671.13
8 10 80 -4.79 22.96 229.96
13 15 195 0.21 0.04 0.65
18 10 180 5.21 27.13 271.26
23 6 138 10.21 104.21 625.26
  48 614     1797.92
Here, N = 48, and $\sum\text{f}_{\text{i}}\text{x}_{\text{i}}=614$
$\overline{\text{x}}=\frac{1}{\text{N}}\big(\sum\text{f}_{\text{i}}\text{x}_{\text{i}}\big)=\frac{614}{48}=12.79$
$\sum\text{f}_{\text{i}}(\text{x}_{\text{i}}-\overline{\text{x}})^2=1797.92$
$\therefore\text{Var}(\text{x})=\frac{1}{\text{N}}\Big[\sum\text{f}_{\text{i}}(\text{x}_{\text{i}}-\overline{\text{x}})^2\Big]=\frac{1797.92}{48}=37.46$
S.D. $=\sqrt{\text{var}(\text{x})}=\sqrt{37.496}=6.12$
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Question 285 Marks
The variance of 15 bservation is 4. If each observation is increased by 9, find the variance of the resulting observation.
Answer
We have, n = 15, and $\sigma^2=4$
Now each observation is increasedd by 9.
Suposs X = x +9 be the new data.
$\therefore\overline{\text{X}}=\frac{1}{15}\sum(\text{x}_\text{i}+9)=\Big(\frac{1}{15}\times\sum\text{x}_\text{i}\Big)+9=\overline{\text{x}}+9$
$\Rightarrow\sum{\text{X}_\text{i}}^{2}=\sum(\text{x}_\text{i}+9)^2=\sum{\text{x}_\text{i}}^2+\sum18\text{x}_\text{i}+\sum9^2$
Since, $\sigma^2=5$
$\Rightarrow\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2=4$
Now, for the new data:
$\sigma^2=\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2=\frac{1}{15}\big(\sum{\text{x}_\text{i}}^2+\sum18\text{x}_\text{i}+\sum9^2\big)-\big(\overline{\text{x}}+9\big)^2$
$=\frac{1}{15}\sum{\text{x}_\text{i}}^2+\frac{1}{15}\sum18\text{x}_\text{i}+\frac{1}{15}\sum9^2-(9)^2-(18\overline{\text{x}})-(\overline{\text{x}})^2$
$=\Big[\frac{1}{15}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2\Big]+\Big[\frac{1}{15}\sum18\text{x}_\text{i}-(18\overline{\text{x}})\Big]+\Big[\frac{1}{15}\sum9^2-(9)^2\Big]$
$=\Big[\frac{1}{15}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2\Big]+\Big[18\times\frac{1}{15}\sum\text{x}_\text{i}-(18\overline{\text{x}})\Big]+\Big[\frac{1}{15}\times15\times(9)^2-(9)^2\Big]$
$=\frac{1}{15}\sum{\text{x}_\text{i}}^2-(\overline{\text{x}})^2$
$=4$
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Question 295 Marks
For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.
Answer
We have,
$n=200, \overline{X}=40, \sigma=15 .$
$\therefore \overline{X}=\frac{1}{n} \sum x_{i}=\overline{X}=200 \times 40=8000 \text {. }$
$\text { Corrected } \sum x_{i}=\text { Incorrected } \sum x_{i}-(\text { sum of incorrect values) }+ \text { (sum of correct values) }$
$=8000-34-53+43+35=7991$
$\therefore \text { Corrected mean }=\frac{\text { corrected } \sum x_1}{n}=\frac{7991}{200}=39.955$
$\text { Now } \sigma=15$
$\Rightarrow 15^2=\frac{1}{200}\left(\sum x_{i}^2\right)-\left(\frac{1}{200} \sum x_{i}\right)^2$
$\Rightarrow 255=\frac{1}{200}\left(\sum x_{i}^2\right)-\left(\frac{8000}{200}\right)^2$
$\Rightarrow 255=\frac{1}{200}\left(\sum x_{i}^2\right)-1600$
$\sum x_{i}^2=200 \times 1825=365000$
$\text { Incorrect } \sum x_{i}^2=365000$
$\text { corrected } \left.\sum x_{i}^2=\text { (incorrected } \sum x_{i}^2\right)-(\text { sum of squers of incorrect values) }+ \text { (sum of squers of correct values) }$
$=365000-(34)^2-53^2+(43)^2+35^2=364109$
$\text { so, Corrected } \sigma=\sqrt{\frac{1}{n} \sum x_{i}^2-\left(\frac{1}{n} \sum x_{i}\right)^2}=\sqrt{\frac{364109}{200}-\left(\frac{7991}{200}\right)^2}$
$=\sqrt{1820.545-1596.402}=14.97$
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