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MATHS 5 Marks Questions

Statistics · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 15 questions.

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15 questions · timed · auto-graded

Question 15 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
Given that, $\text{n}=100,\ \bar{\text{x}}=40,\ \sigma=10$
$\therefore\ \bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{N}}$
$\Rightarrow40=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow{\sum\text{x}_\text{i}}=4000$
Corrected ${\sum\text{x}_\text{i}}=4000-30-70+3+27=3930$
and Corrected mean $=\frac{3930}{100}=39.3$
Now, $\sigma^2=\frac{\sum\text{x}_\text{i}^2}{\text{n}}-(40)^2$
$\Rightarrow100=\frac{\sum\text{x}_\text{i}^2}{100}-1600$
$\Rightarrow{\sum\text{x}_\text{i}^2}=1700\times100$
$\Rightarrow{\sum\text{x}_\text{i}^2}=170000$
$\therefore\text{ Corrected }{\sum\text{x}_\text{i}^2}=170000-(30)^2-(70)^2+(3)^2+(27)^2$
 $=170000-900-4900+9+729=164938$
$\therefore\ \text{Correct SD}=\sqrt{\frac{164938}{100}-(39.3)^2}$
$=\sqrt{1649.38-1544.49}$
$=\sqrt{104.89}=10.24$
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Question 25 Marks
Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.
Answer
First n natural numbers are 1, 2, 3, ........, n. Here, n is odd.
$\therefore\text{ Mean }\bar{\text{x}}=\frac{1+2+3+\ ......\ +\text{n}}{\text{n}}=\frac{\frac{\text{n}(\text{n}+1)}{2}}{2}=\frac{\text{n}+1}{2}$
The deviations of numbers from mean $\Big(\frac{\text{n}+1}{2}\Big)$ are
$1-\frac{\text{n}+1}{2},2-\frac{\text{n}+1}{2},3-\frac{\text{n}+1}{2},\ .......\ ,\text{n}-\frac{\text{n}+1}{2}$
$\text{i.e.},-\frac{\text{n}-1}{2},\frac{\text{n}-3}{2},\ ......\ ,-2,-1,0,1,2,\ .....\ ,\frac{\text{n}-1}{2}$
The absolute values of deviation from the mean i.e., $|\text{x}_\text{i}-\bar{\text{x}}|$
$\frac{\text{n}-1}{2},\frac{\text{n}-3}{2},\ ......\ ,2,1,0,1,2,\ .....\ ,\frac{\text{n}-1}{2}$
The sum of absolute values of deviations from the mean i.e. $|\text{x}_\text{i}-\bar{\text{x}}|$
$=2\Big(1+2+3+\ ....\text{ to }\frac{\text{n}-1}{2}\text{terms}\Big)$
$=2\cdot\frac{\frac{\text{n}-1}{2}\big(\frac{\text{n}-1}{2}+1\big)}{2}=\frac{\text{n}-1}{2}\cdot\frac{\text{n}+1}{2}=\frac{\text{n}^2-1}{4}$
$\therefore$ Mean deviation about the mean.
$=\frac{\sum|\text{x}_\text{i}-\bar{\text{x}}|}{\text{n}}=\frac{\frac{\text{n}^2-1}{4}}{\text{n}}=\frac{\text{n}^2-1}{4\text{n}}$
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Question 35 Marks
Find the standard deviation of the first n natural numbers.
Answer
xi
1
2
3
4
5
-
-
n
xi2
1
4
9
16
25
-
-
n2
Solution:
$\sum\text{x}_\text{i}=1+2+3+4+5+\ ....\ +\text{n}=\frac{\text{n}(\text{n}+1)}{2}$
$\sum\text{x}_\text{i}^2=1^2+2^2+3^2+\ ....\ +\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\therefore\ \text{S.D.}(\sigma)=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2}$
$=\sqrt{\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6\text{n}}-\frac{\text{n}^2(\text{n}+1)}{4\text{n}^2}}$
$=\sqrt{\frac{(\text{n}+1)(2\text{n}+1)}{6}-\frac{(\text{n+1}^2)}{4}}$
$=\sqrt{\frac{2\text{n}^2+3\text{n}+1}{6}-\frac{\text{n}^2+2\text{n}+1}{4}}$
$=\sqrt{\frac{4\text{n}^2+6\text{n}+2-3\text{n}^2-6\text{n}-3}{12}}$
$=\sqrt{\frac{\text{n}^2-1}{12}}$
Hence, the required SD $=\sqrt{\frac{\text{n}^2-1}{12}}$
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Question 45 Marks
Calculate the mean deviation from the median of the following data:
Class interval
0-6
6-12
12-18
18-24
24-30
Frequency
4
5
3
6
2
Answer
Class interval
fi
Xi
c.f.
$\text{d}_\text{i}=|\text{x}_\text{i}-{\text{Med}}|$
fidi
0-6
4
3
4
11
44
6-12
5
9
9
5
25
12-18
3
15
12
1
3
18-24
6
21
18
7
42
24-30
2
27
20
13
26
 
N = 20
 
 
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=140$
$\text{Median class}=\Big(\frac{\text{N}}{2}\Big)^{\text{th}}\text{ term}=\frac{20}{2}^{\text{th}}\text{ term}$
$=10^{\text{th}}\text{ term i.e. }12-18$
$\therefore\ \text{Median}=\text{l}+\frac{\frac{\text{N}}{2}-\text{cf}}{\text{f}}\times\text{h}$
$=12+\frac{10-9}{3}\times6=12+\frac{1}{3}\times6=12+2=14$
and $\text{MD}=\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\text{N}}=\frac{140}{20}=7$
Hence, the required MD = 7
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Question 55 Marks
There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
Marks
0
1
2
3
4
5
Frequency
x - 2
x
x2
(x + 1)2
2x
x + 1
where x is a positive integer. Determine the mean and standard deviation of the marks.
Answer
Sum of frequencies,
x - 2 + x + x2 + (x + 1)2 + 2x + x + 1 = 60 (given)
⇒ 2x2 + 7x - 60 = 0
⇒ (2x + 15)(x - 4) = 0
⇒ x = 4
xi
fi
di = xi - 3
fidi
fidi2
0
2
-3
-6
18
1
4
-2
-8
16
2
16
-1
-16
16
A = 3
25
0
0
0
4
8
1
8
8
5
5
2
10
20
Total
$\sum\text{f}_\text{i}=60$
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=-12$
 $\sum\text{f}_\text{i}\text{d}_\text{i}^2=78$
$\text{Mean}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}=3+\Big(\frac{-12}{60}\Big)=2.8$
Standard Deviation,
$\sigma=\sqrt{\frac{\sum\text{f}_\text{i}\text{d}^2_\text{i}}{\sum\text{f}_\text{i}}-\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}\Big)^2}$
$=\sqrt{\frac{78}{60}-\Big(\frac{-12}{60}\Big)^2}$
$=\sqrt{1.3-0.04}$
$=\sqrt{1.26}=1.12$
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Question 65 Marks
The mean and standard deviation of a set of n1 observations are $\bar{\text{x}}_1$ and S1, respectively while the mean and standard deviation of another set of n2 observations are $\bar{\text{x}}_2$ and S2, respectively. Show that the standard deviation of the combined set of (n1 + n2 ) observations is given by $\text{S.D.}=\sqrt{\frac{\text{n}_1(\text{s}_1)^2+\text{n}_2(\text{s}_2)^2}{\text{n}_1+\text{n}_2}+\frac{\text{n}_1\text{n}_2(\bar{\text{x}}_1-\bar{\text{x}}_2)^2}{(\text{n}_1+\text{n}_2)^2}}$
Answer
We have two sets of observations,
$\text{x}_\text{i},\text{ i}=1,2,3,\ .....\ ,\text{n}_1$ and $\text{y}_\text{j},\text{ j}=1,2,3,\ ....,\ \text{n}_2$
$\therefore\ \bar{\text{x}}_1=\frac{1}{\text{n}_\text{1}}\sum\limits^{\text{n}_1}_{\text{i}=1}\text{x}_\text{i}$ and $\bar{\text{x}}_2=\frac{1}{\text{n}_\text{2}}\sum\limits^{\text{n}_2}_{\text{j}=1}\text{y}_\text{j}$
$\Rightarrow\ \sigma^2_1=\frac{1}{\text{n}_1}\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}}_1)^2$ and $\sigma^2_2=\frac{1}{\text{n}_2}\sum\limits^{\text{n}_2}_{\text{j}=1}(\text{y}_\text{j}-\bar{\text{x}}_2)^2$
Now, mean $\bar{\text{x}}$ of the given series is given by
$\bar{\text{x}}=\frac{1}{\text{n}_1+\text{n}_2}\bigg[\sum\limits^{\text{n}_1}_{\text{i}=1}\text{x}_\text{i}+\sum\limits^{\text{n}_2}_{\text{j}=1}\text{y}_\text{j}\bigg]=\frac{\text{n}_1\bar{\text{x}}_1+\text{n}_2\bar{\text{x}}_2}{\text{n}_1+\text{n}_2}$
The variance $\sigma^2$ of the combined series is given by,
$\sigma^2=\frac{1}{\text{n}_1+\text{n}_2}\bigg[\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_{\text{i}}-\bar{\text{x}})^2+\sum\limits^{\text{n}_2}_{\text{j}=1}(\text{y}_\text{j}-\bar{\text{x}})^2\bigg]$
Now, $\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2=\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}}_\text{j}+\bar{\text{x}}_\text{j}-\bar{\text{x}})^2$
But $\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})=0$ [Algebraic sum of the deviation of values of first series from mean is zero]
Also, $\sum\limits^{\text{n}_1}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2=\text{n}_1\text{s}^2_1=\text{n}_1(\bar{\text{x}}_1-\bar{\text{x}})^2=\text{n}_1\text{s}^2_1+\text{n}_1\text{d}^2_1$
Where, $\text{d}_1=(\bar{\text{x}}_1-\bar{\text{x}})$
Similarly, $\sum\limits^{\text{n}_2}_{\text{j}=1}(\text{y}_\text{j}-\bar{\text{x}})^2=\sum\limits^{\text{n}_2}_{\text{j}=1}(\text{y}_\text{j}-\bar{\text{x}}_\text{i}+\bar{\text{x}}_\text{i}-\bar{\text{x}})^2=\text{n}_2\text{s}^2_2+\text{n}_2\text{d}^2_2$
Where, $\text{d}_2=\bar{\text{x}}_2-\bar{\text{x}}$
Combined SD, $\sigma=\sqrt{\frac{\big[\text{n}_1\big(\text{s}^2_1+\text{d}^2_1\big)+\text{n}_2\big(\text{s}^2_2+\text{d}^2_2\big)\big]}{\text{n}_1+\text{n}_2}}$
Where, $\text{d}_1=\bar{\text{x}}_1-\bar{\text{x}}=\bar{\text{x}}_1-\Big(\frac{\text{n}_1\bar{\text{x}}_1+\text{n}_2\bar{\text{x}_2}}{\text{n}_1+\text{n}_2}\Big)=\frac{\text{n}_2(\bar{\text{x}}_1-\bar{\text{x}}_2)}{\text{n}_1+\text{n}_2}$
and $\text{d}_2=\bar{\text{x}}_2-\bar{\text{x}}=\bar{\text{x}}_2-\frac{\text{n}_1\bar{\text{x}}_1+\text{n}_2\bar{\text{x}_2}}{\text{n}_1+\text{n}_2}=\frac{\text{n}_1(\bar{\text{x}}_2-\bar{\text{x}}_1)}{\text{n}_1+\text{n}_2}$
 $\therefore\ \sigma^2=\frac{1}{\text{n}_1+\text{n}_2}\bigg[\text{n}_1\text{s}^2_1+\text{n}_2\text{s}^2_2+\frac{\text{n}_1\text{n}_2(\bar{\text{x}}_1+\bar{\text{x}}_2)^2}{(\text{n}_1+\text{n}_2)^2}+\frac{\text{n}_2\text{n}_1(\bar{\text{x}}_2+\bar{\text{x}}_1)^2}{(\text{n}_1+\text{n}_2)^2}\bigg]$
Also, $\sigma=\sqrt{\frac{\text{n}_1\text{s}^2_1+\text{n}_2\text{s}^2_2}{\text{n}_1+\text{n}_2}+\frac{\text{n}_1\text{n}_2(\bar{\text{x}}_1-\bar{\text{x}}_2)^2}{(\text{n}_1+\text{n}_2)^2}}$
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Question 75 Marks
The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.
Answer
Given that $\text{n}_1=60,\ \bar{\text{x}}_1=650,\ \text{s}_1=8$
And $\text{n}_2=80,\ \bar{\text{x}}_2=660,\ \text{s}=7$
We know that for a combined series.
$\sigma=\sqrt{\frac{\text{n}_1\text{s}^2_1+\text{n}_2\text{s}^2_2}{\text{n}_1+\text{n}_2}+\frac{\text{n}_1\text{n}_2(\bar{\text{x}}_1-\bar{\text{x}}_2)^2}{(\text{n}_1+\text{n}_2)^2}}$
$=\sqrt{\frac{60\times(8)^2+80\times(7)^2}{60+80}+\frac{60\times80(650-660)^2}{(60+80)^2}}$
$=\sqrt{\frac{6\times64+8+49}{14}+\frac{60\times80\times100}{140\times140}}$
$=\sqrt{\frac{192+196}{7}+\frac{1200}{49}}=\sqrt{\frac{338}{7}+\frac{1200}{49}}$
$=\sqrt{\frac{2716+1200}{49}}=\sqrt{\frac{3916}{49}}$
$=\frac{62.58}{7}=8.9$
Hence, the required SD = 8.9
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Question 85 Marks
Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.
Ravi
25
50
45
30
72
42
36
48
35
60
Hashima
10
70
50
20
95
55
42
60
48
80
Who is more intelligent and who is more consistent?
Answer
For Ravi,
xi
di = xi - 45
di2
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
Total
$\sum\text{d}_\text{i}=-9$
$\sum\text{d}^2_\text{i}=1699$
$\sigma=\sqrt{\frac{{\sum\text{d}_\text{i}}}{\text{n}}-\Big(\frac{\sum\text{d}_\text{i}}{\text{n}}\Big)^2}=\sqrt{\frac{1699}{10}-\Big(\frac{-9}{10}\Big)^2}$
$=\sqrt{169.9-0.81}=\sqrt{169.09}=13.003$
Now, $\bar{\text{x}}=\text{A}+\frac{\sum\text{d}_\text{i}}{\sum\text{f}_\text{i}}=45-\frac{14}{10}=43.6$
For Hashina,
xi
di = xi - 55
di2
10
-45
2025
70
15
225
50
-5
25
20
-35
1225
95
40
1600
55
0
0
42
-13
169
60
5
25
48
-7
49
80
25
625
Total
$\sum\text{d}_\text{i}=-20$
$\sum\text{d}^2_\text{i}=5968$
$\therefore\ \sigma=\sqrt{\frac{5968}{10}-\Big(\frac{-20}{10}\Big)^2}$
$=\sqrt{596.8-4}=\sqrt{592.8}=24.46$
For Ravi, $\text{CV}=\frac{\sigma}{\bar{\text{x}}}\times100=\frac{13.003}{43.6}\times100=29.82$
For Hashina, $\text{CV}=\frac{\sigma}{\bar{\text{x}}}\times100=\frac{24.46}{55}\times100=44.47$
Hence, Hashina is more consistent and intelligent.
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Question 95 Marks
The frequency distribution:
x
A
2A
3A
4A
5A
6A
f
2
1
1
1
1
1
where A is a positive integer, has a variance of 160. Determine the value of A.
Answer
x
fi
fixi
fixi2
A
2
2A
2A2
2A
1
2A
4A2
3A
1
3A
9A2
4A
1
4A
16A2
5A
1
5A
25A2
6A
1
6A
36A2 
Total
n = 7
$\sum\text{f}_\text{i}\text{x}_\text{i}=22\text{A}$
$\sum\text{f}_\text{i}\text{x}_\text{i}^2=92\text{A}^2$
$\therefore\ \sigma^2=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{n}}\Big)^2$
$\Rightarrow160=\frac{92\text{A}^2}{7}-\Big(\frac{224}{7}\Big)^2$
$\Rightarrow160=\frac{92\text{A}^2}{7}-\frac{484\text{A}^2}{49}$
$\Rightarrow160=(644-484)\frac{\text{A}^2}{49}$
$\Rightarrow160=\frac{160\text{A}^2}{49}$
$\Rightarrow\text{A}^2=49\ \therefore\ \text{A}=7$
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Question 105 Marks
Calculate the mean deviation about the mean for the following frequency distribution:
Class interval
0-4
4-8
8-12
12-16
16-20
Frequency
4
6
8
5
2
Answer
Class interval
fi
Xi
fixi
$\text{d}_\text{i}=|\text{x}_\text{i}-\bar{\text{x}}|$
fidi
0-4
4
2
8
7.2
28.8
4-8
6
6
36
3.2
19.2
8-12
8
10
80
0.8
6.4
12-16
5
14
70
4.8
24.0
16-20
2
18
36
8.8
17.6
 
N = 16
 
$\sum\text{f}_\text{i}\text{x}_\text{i}=230$
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=96.0$
$\text{Mean }=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}=\frac{230}{25}=9.2$
And Mean $\text{deviation}=\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\text{N}}=\frac{96}{25}=3.84$
Hence, the required MD = 3.84
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Question 115 Marks
Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.
Answer
xi
xi - a
(xi - a)2
a
0
0
a + d
d
d2
a + 2d
2d
4d2
....
....
9d2
....
....
....
....
....
....
a + (n - 1)d
(n - 1)d
(n - 1)2d2
$\sum\text{x}_\text{i}=\frac{\text{n}}{2}\big[2\text{a}+(\text{n}-1)\text{d}\big]$
 
  
$\because\ \text{Mean}=\frac{\sum\text{x}_\text{i}}{\text{n}}=\frac{1}{\text{n}}\Big[\frac{\text{n}}{2}\big(2\text{a}+(\text{n}-1)\big)\text{d}\Big]=\text{a}+\frac{(\text{n}-1)}{2}\text{d}$
$\therefore\ \sum(\text{x}_\text{i}-\text{a})=\text{d}\big[1+2+3+\ .....\ +(\text{n}-1)\text{d}\big]=\text{d}+\frac{(\text{n}-1)\text{n}}{2}$
and $\sum(\text{x}_\text{i}-\text{a})^2=\text{d}^2\big[1^2+2^2+3^2+\ .....\ +(\text{n}-1)^2\text{d}\big]=\frac{\text{d}^2\text{n}(\text{n}-1)(2\text{n}-1)}{6}$
$\sigma=\sqrt{\frac{(\text{x}_\text{i}-\text{a})^2}{\text{n}}-\Big(\frac{\text{x}_\text{i}-\text{a}}{\text{n}}\Big)^2}$
$=\sqrt{\frac{\text{d}^2(\text{n})(\text{n}-1)(2\text{n}-1)}{6\text{n}}-\Big[\frac{\text{d}(\text{n}-1)\text{n}}{2\text{n}}\Big]^2}$
$=\sqrt{\frac{\text{d}^2(\text{n}-1)(2\text{n}-1)}{6}-\frac{\text{d}^2(\text{n}-1)^2}{4}}$
$=\text{d}\sqrt{\frac{(\text{n}-1)(\text{n}+1)}{12}}$
$=\text{d}\sqrt{\frac{\text{n}^2-1}{12}}$
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Question 125 Marks
Determine the mean and standard deviation for the following distribution:
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
Marks
fi
fixi
$\text{d}_\text{i}=\text{x}_\text{i}-\bar{\text{x}}$
fidi
fidi2
2
1
2
-4
-4
16
3
6
18
-3
-18
54
4
6
24
-2
-12
24
5
8
40
-1
-8
8
6
8
40
-1
-8
8
7
2
14
1
2
2
8
2
16
2
4
8
9
3
27
3
9
27
10
0
0
4
0
0
11
2
22
5
10
50
12
1
12
6
6
36
13
0
0
7
0
0
14
0
0
8
0
0
15
0
0
9
0
0
16
1
16
10
10
100
Total
$\sum\text{f}_\text{i}=40$
$\sum\text{f}_\text{i}\text{x}_\text{i}=239$
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=-1$
$\sum\text{f}_\text{i}\text{x}^2_\text{i}=325$
$\therefore\ \text{Mean }\bar{\text{x}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{239}{40}=5.975\approx6$
and $\sigma=\sqrt{\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}-\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}\Big)^2}$
$=\sqrt{\frac{325}{40}-\Big(\frac{-1}{40}\Big)^2}$
$=\sqrt{8.125-0.000625}$
$=\sqrt{8.124375}$
$=2.85$
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Question 135 Marks
While calculating the mean and variance of 10 readings, a student wrongly used the reading 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
Given, $\text{n}=10,\ \bar{\text{x}}=45,\ \sigma^2=16$
$\bar{\text{x}}=45$
$\Rightarrow\frac{\sum\text{x}_\text{i}}{\text{n}}=45$
$\Rightarrow{\sum\text{x}_\text{i}}=450$
Corrected ${\sum\text{x}_\text{i}}=450-52+25=423$
$\therefore$ Corrected mean, $\bar{\text{x}}=\frac{432}{10}=42.3$ 
 $\Rightarrow\ \sigma^2=\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2$
$\Rightarrow16=\frac{\sum\text{x}_\text{i}^2}{10}-(45)^2$
$\Rightarrow{\sum\text{x}_\text{i}^2}=20410$
$\therefore\text{ Corrected }{\sum\text{x}_\text{i}^2}=20410-(53)^2+(25)^2=18331$
And $\text{Corrected }\sigma^2=\frac{18331}{10}-(42.3)^2=43.81$
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Question 145 Marks
The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results: Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds. Further, another set of 15 observations x1, x2, ..., x15, also in seconds, is now available and we have $\sum\limits^{15}_{\text{i}=1}\text{x}_\text{i}=279$ and $\sum\limits^{15}_{\text{i}=1}\text{x}_\text{i}^2=5524.$ Calculate the standard derivation based on all 40 observations.
Answer
Given $\text{n}_1=25,\ \bar{\text{x}}_\text{i}=18.2,\ \sigma_1=3.25$
$\text{n}_2=15,\ \sum\limits^{15}_{\text{i}=1}\text{x}_\text{i}=279$ and $\sum\limits^{15}_{\text{i}=1}\text{x}_\text{i}^2=5524$
For first set $\sum\text{x}_\text{i}=25\times18.2=455$
$\therefore\ \sigma^2_1=\frac{\sum\text{x}^2_\text{i}}{25}-(18.2)^2$
 $\Rightarrow(3.25)^2=\frac{\sum\text{x}^2_\text{i}}{25}-(18.2)^32$
$\Rightarrow10.5625+331.24=\frac{\sum\text{x}^2_\text{i}}{25}$
$\Rightarrow\sum\text{x}^2_\text{i}=25\times(10.5625+331.24)$
$\Rightarrow\sum\text{x}^2_\text{i}=25\times341.8025=8545.0625$
For combined SD of the 40 observations, n = 40
Now, $\sum\limits^{40}_{\text{i}=1}\text{x}^2_\text{i}=5524+8545.0625=14069.0625$
and $\sum\limits^{40}_{\text{i}=1}\text{x}_\text{i}=455+279=734$
$\therefore\ \text{SD}=\sqrt{\frac{14069.0625}{40}-\Big(\frac{734}{40}\Big)^2}$
$=\sqrt{351.1726-(18.35)^2}$
$=\sqrt{351.726-336.7225}$
$=\sqrt{15.0035}$
$=3.87$
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Question 155 Marks
The weights of coffee in 70 jars is shown in the following table:
Weight (in grams)
Frequency
200-201
13
201-202
27
202-203
18
203-204
10
204-205
1
205-206
1
Determine variance and standard deviation of the above distribution.
Answer
Weight (in grams)
Frequency
xi
di = xi - A
fidi
fidi2
200-201
13
200.5
-2
-26
52
201-202
27
201.5
-1
-27
27
202-203
18
202.5(A)
0
0
0
203-204
10
203.5
1
10
10
204-205
1
204.5
2
2
4
205-206
1
205.5
3
3
9
 
N = 70
 
 
$\sum\text{f}_\text{i}\text{d}_\text{i}=-38$
$\sum\text{f}_\text{i}\text{d}_\text{i}^2=102$
$\therefore\ \text{Variance}=\sigma^2=\frac{\sum\text{f}_\text{i}\text{d}^2_\text{i}}{\text{N}}-\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\text{N}}\Big)^2$
$=\frac{102}{70}-\Big(\frac{-38}{70}\Big)^2=1.457-0.292=1.165$
$\therefore\ \text{SD}=\sigma=\sqrt{1.165}=1.08\text{g}$
Hence, the required variance = 1.165 and SD = 1.08g
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