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Applied Maths 2 Marks Questions

Inferential Statistics · CBSE STD 12 Science (CBSE - English Medium). 15 questions.

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2 Marks Questions

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Question 12 Marks
Answer
(1)
We have $\mu=$ population mean $=100$
n = sample size = 10
computation of $\bar{X}$ and S
$x_i$$d_i=x_i-90$$d_1^2$
70-20400
12030900
10020400
10111121
88-24
83-749
95525
98864
10717289
10010100
$\sum d_i=72$$\sum d_i^2=2352$
$\therefore \quad \bar{X}=90+\frac{1}{10} \sum d_i$
$\left[\right.$ Using $\left.\bar{X}=A+\frac{1}{n} \sum d_i\right]$
$=90+\frac{72}{10}=97.2$
and $S^2=\frac{1}{n-1}\left\{\sum d_i^2-\frac{1}{n}\left(\sum d_i\right)^2\right\}$
$\begin{array}{l}=\frac{1}{9}\left\{2352-\frac{(72)^2}{10}\right\} \\ =\frac{1833.6}{9}=203.73\end{array}$
Now, $t=\frac{\bar{X}-\mu}{\frac{S}{\sqrt{n}}}$
$=\frac{97.2-100}{\sqrt{\frac{203.73}{10}}}$
$\begin{array}{l}=\frac{-2.8}{\sqrt{20.37}} \\ =\frac{-2.8}{4.514}=-0.62\end{array}$
$\Rightarrow \quad|t|=0.62$

(2)
We define,
Null Hypothesis $H _0$ :The data are consistent with the assumption of a mean L.Q. of 100 in the population.
Alternate Hypothesis $H _1$ : The mean I.Q. of population + 100.
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Question 22 Marks
Read the following text and answer the following questions on the basis of the same: Each sub-part carries 2 marks
are found to be 100, 104, 108, 110, 118, 120, 122, 124, 126, and 128 cms.

Image

Q.1. Find the t-statistic for the given data, if mean height of the students is 110 cms.
Q.2. Discuss the suggestion that the mean height of the students of the college is 110 cms.[Given $t_9(0.05)=$2.262]
Answer

(1)
Let the sample statistic 1 be given by
$t=\frac{\bar{X}-\mu}{\frac{S}{\sqrt{n}}}$
Let us now compute the sample mean $(\bar{X})$ and S .
Computation of $\bar{X}$ and $S$

$x_i$$x_i-\bar{X}$$\left(x_i-\bar{X}\right)^2$
100-16256
104-12144
108-864
110-636
11824
120416
122636
124864
12610100
12812144
$\sum x_i=1160$ $\sum_{i=1}^{10}\left(x_i-\bar{X}\right)^2=864$


Here
$\bar{X}=\frac{\sum x_i}{10}=\frac{1160}{10}=116$
$\therefore \quad t=\frac{\bar{X}-\mu}{\frac{S}{\sqrt{n}}}$
$\Rightarrow \quad t=\frac{116-110}{\frac{9.798}{\sqrt{10}}}$
$=\frac{6}{9.798} \times 3.162 \quad[\because \mu=110]$
= 1.94
$\left[\because S^2=\frac{1}{n-1} \sum_{i=1}^{10}\left(x_i-\bar{X}\right)^2 \Rightarrow S^2=\frac{1}{9} \times 864\right.$
$\left.\Rightarrow S=\frac{\sqrt{864}}{3}=\frac{29.393}{3}=9.798\right]$

(2)
We define
Null Hypothesis $H _0$ : There is no significant difference between the sample mean and hypothetical population mean 110 cms.
Alternate Hypothesis is $H _1$ : The sample mean is not same as the population mean. 1 Here, the sample statistic follows student's 1distribution -with v = (10 - 1) = 9 degree of freedom.
We shall now compare the calculated t-test value with the value of f for 9 degrees of freedom at a certain level of significance. It is given that $t_9(0.05)$= 2.262
$\therefore \quad$ Calculated $|t|=1.94<2.262=t_9(0.05)$
i.c., Calculated $|t|<$ tabulated $t_9(0.05)$
So, we accept the null hypothesis. Hence, the sample mean is same as the population mean. Consequently, the mean height of the students of the school is 110 cms.

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Question 32 Marks
3. A machine which produces mica insulating washers of use in the electric devices is set to turn out washers having a thickness of 10 mils ( 1mil = 0.001 inch). A sample of 10 washers has an average thickness of 9.52 mils with a standard deviation of 0.60 mil. Find out t
Answer
$$ Here, $\bar{X}=9.52$, Mean, $\mu=10, \sigma=0.60$ and $n=10$
$\begin{aligned} t & =\frac{Z}{\text { S.E. }} \\ & =\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}} \\ & =\frac{9.52-10}{\frac{0.60}{\sqrt{10}}} \\ & =\frac{-0.48}{\frac{0.60}{3.162}}\end{aligned}$
$\begin{array}{l}=\frac{-0.48}{0.1897} \\ =-2.5303\end{array}$
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Question 42 Marks
In a sample of 400 population from a village 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. Compute the standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village ?
Answer
Given sample size 400 and 230 are vegetarian eaters. So sample proportional $=\frac{230}{400}=0.575$
Population proportion P = Prob. (vegetarian eaters from the village) $=\frac{1}{2}$
(Since vegetarian and non-vegetarian foods are equally popular),
$\begin{aligned} Q & =1-P \\ & =1-\frac{1}{2} \\ & =\frac{1}{2}\end{aligned}$
The standard error S.E =
$\begin{array}{l}=\sqrt{\frac{P Q}{N}} \\ =\sqrt{\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}{400}} \\ =\sqrt{\frac{0.25}{400}} \\ =\sqrt{0.000625} \\ =0.025\end{array}$
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Question 52 Marks
A wholesaler in apples claims that only 4% of the apples supplied by him are defective. A random sample of 600 apples contained 36 defective apples. Calculate the standard error concerning good apples.
Answer
Sample size = 600
No. of defective apples = 36
Sample proportion $P=\frac{36}{600}=0.06$
Population proportion P = probability of defective apples = 4% = 0.04
Q = 1 - P
= 1-0.04= 0.96
The S.E. for sample proportion is given by S.E.
$\begin{array}{l}=\sqrt{\frac{P Q}{N}} \\ =\sqrt{\frac{(0.04)(0.96)}{600}} \\ =\sqrt{0.000064}=0.008\end{array}$
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Question 62 Marks
A sample of 1000 students whose mean weight is 119 lbs (pounds) from a school in Uttar Pradesh State was taken and their average weight was found to be 120 lbs with a standard deviation of 30 Ibs. Calculate the standard error of the mean.
Answer
Given $n=1000$, mean $=119, \sigma=30$
$\begin{aligned} \text { S.E. } & =\frac{\sigma}{\sqrt{n}} \\ & =\frac{30}{\sqrt{1000}} \\ & =\frac{30}{31.623} \\ & =0.9487\end{aligned}$
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Question 72 Marks
A random sample of 60 observations was drawn from a large population and its standard deviation was found to be 2.5. Calculate the suitable standard error that this sample is taken from a population with standard deviation 3?
Answer
Given sample size n = 60
Sample standard deviation = 2.5
Population standard deviation $\sigma=3$
The S.E. is given by
$\begin{aligned} \sqrt{\frac{\sigma^2}{2 n}} & =\sqrt{\frac{9}{120}} \\ & =\sqrt{0.075} \\ & =0.2739\end{aligned}$
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Question 82 Marks
A sample of 100 students is chosen from a large group of students. The average height of these students is 162 cm and standard deviation (S.D.) is 8 cm. Obtain the standard error for the average height of large group of students of 160 cm?
Answer
Given n = 100 $\bar{x}=162 cm, s =8 cm$ is known in this problem
Since $\sigma$ is unknown, so we consider $\hat{\sigma}=s$ and $\varphi=160 cm$
$S . E .=\frac{\hat{\sigma}}{\sqrt{n}}$
$\begin{array}{l}=\frac{s}{\sqrt{n}} \\ =\frac{8}{\sqrt{100}} \\ =0.8\end{array}$
Therefore the standard error for the average height of large group of students of 160 cm is 0.8.
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Question 92 Marks
The standard deviation of a sample of size 50 is 6.3. Determine the standard error whose population standard deviation is 6?
Answer
Sample size $n=50$
Sample S.D. $s=6.3$
Population S.D. $\sigma=6$
The standard error for sample S.D. is given by
$\begin{aligned} \text { S.E. } & =\sqrt{\frac{\sigma^2}{2 n}} \\ & =\frac{6}{\sqrt{2(50)}} \\ & =\frac{6}{\sqrt{100}} \\ & =0.6\end{aligned}$
Thus standard error for sample S. D = 0.6 .
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Question 102 Marks
Find the sample size for the given standard deviation 10 and the standard error with respect of sample mean is 3.
Answer
Given $s=10$, S.E. $\bar{X}=3$
Since,$\text { S.E. }=\frac{\sigma}{\sqrt{n}}$
Therefore,$3=\frac{10}{\sqrt{n}}$
$\Rightarrow \quad \sqrt{n}=\frac{10}{3}$
Taking square on both sides we get
$n=\left(\frac{10}{3}\right)^2$
$=\frac{100}{9}$
$\begin{array}{l}=11.11 \\ \cong 11,\end{array}$
The required sample size is 11 .
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Question 112 Marks
A server channel monitored for an hour was found a have an estimated mean of 20 transactions transmitted per minute. The variance is known to be 4. Find the standard error.
Answer
Given $\sigma^2=4$ which implies $\sigma=2, n=1$ hour $=60$, $\min , \bar{X}=20 / min$
Standard Error $=\frac{\sigma}{\sqrt{n}}$
$\begin{array}{l}=\frac{2}{\sqrt{60}} \\ =0.2582\end{array}$
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Question 122 Marks
Define parameter in Statistics.
Answer
A parameter is any numerical quantity that characterizes a given population or some aspect of it. This means the parameter tells us something about the whole population. For example, the population mean $\mu$, variance $\sigma^2$, population proportion P, population correlation p.
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Question 132 Marks
What is statistic?
Answer
A statistic is used to estimate the value of a population parameter. For instance, we selected a random sample of 100 students from a school with 1000 students. The average height of the sampled students would be an example of a statistic.
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Question 142 Marks
What is sample ?
Answer
A sample is a set of data collected from a statistical population by a defined procedure. The elements of a sample are called sample size or sample. Samples are collected and statistics are calculated from the samples, so that one can make inferences from the sample to the population.
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Question 152 Marks
What is population?
Answer
A population is a set of similar items or events which is of interest for some question or experiment. A population can be specific or vague. Examples of population defined vaguely include the number of newborn babies in Uttar Pradesh, a total number of tech startups in India, the average height of all exam candidates, mean weight of taxpayers in Lucknow etc. Examples of population defined specifically include a number of fans produced in a particular factory, the number of students in a class, the number of boys and girls in a coaching centre etc.
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2 Marks Questions - Applied Maths STD 12 Science Questions - Vidyadip