Question

Professor Smith took a sample of 196 students from her morning class and a sample of 171 students from her evening class. On Exam 1, all those sampled from the morning class got a 66 while those sampled from the evening class averaged a 67 with a variance of 178. If Professor Smith combines these two samples. Question 2 What is the combined Exam 1 average? Question 3AWhat is the combined Exam 1 median? Question 3B What is the shape of the combined Exam 1 grade distribution? Question 4 What is the combincd Exam 1 standard deviation?

Answer 1

Let the averages of the sample from the morning class and the evening class be  $\bar{x_{1}}$ and $\bar{x_{2}}$ respectively.

Let the variances of the sample from the morning class and the evening class be  $\sigma_{1}^2$ and $\sigma_{2}^2$​​​​ respectively.

Let the sizes of the sample from the morning class and the evening class be n₁ and n₂ respectively.

Given:- n₁=196 , n₂=171 , $\bar{x_{2}}$ =67 , $\sigma_{2}^2$ =178

Since the sample from the morning class all got the same marks 66, it is clear that $\bar{x_{1}}$ =66 , $\sigma_{1}^2$ =0

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Question 2

$combined\; \;average=\bar{x} =\frac{n_{1}\bar{x_{1}}+n_{2}\bar{x_{2}}}{n_{1}+n_{2}} =\frac{196\times66+171\times67}{196+171}=66.466$

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Question 3A

Total number of observations = 196+171 = 367 ; which is a odd number.

So, after arranging the marks of 367 students in increasing order, combined median is the 184th observation. [ $\because$ (367+1)/2=184]

In the sample from the evening class, the average marks is 67. This implies that there are some students among the 171 students who got less marks than 66 and there are some students among the 171 students who got more marks than 66.

Let the number of students from the sample from the evening class who got more marks than 66 be m [0<m<171].

Thus the number of students from the sample from the evening class who got marks less than or equal to 66 = 171-m < 171

Therefore in the combined sample, after arranging all the students with marks in increasing order;

• the first (171-m) students have marks less than or equal to 66.
• the next 196 students all have marks equal to 66.
• the last m students have marks more than 66.

Therefore in the combined sample, total number of students who have marks less than or equal to 66 = 171-m+196 = 367-m

Now; m<171 $\Rightarrow$ 367-m>367-171 $\Rightarrow$ 367-m>196

Therefore in the combined sample, after arranging all the students with marks in increasing order; the 184th student has marks less than or equal to 68.  

Also; 171-m<171 $\Rightarrow$ 171-m$\leqslant$170

Therefore in the combined sample, the first 170 students have marks less than or equal to 66; which implies that the 184th student has marks equal to 66.

$\therefore$COMBINED MEDIAN = 66  

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Question 3B

There are at least 196 students with marks equal to 66, which implies that at most (367-196)=171 students have got marks other than 66. [171<196]

Therefore, most of the students have got marks equal to 66 $\Rightarrow$mode of the combined sample = 66

Therefore the combined average, the combined median and the combined have almost the same value; which is a feature of a symmetric distribution.

$\therefore$  The shape of the combined distribution is mound shape symmetrical.

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Question 4

$combined\; \;variance=\frac{n_{1}\sigma_{1}^2&plus;n_{2}\sigma_{1}^2}{n_{1}&plus;n_{2}}&plus;\frac{n_{1}(\bar{x_{1}}-\bar{x})^2&plus;n_{2}(\bar{x_{2}}-\bar{x})^2}{n_{1}&plus;n_{2}}$

$=\frac{196\times0&plus;171\times178}{196&plus;171}&plus;\frac{196(66-66.466)^2&plus;171(67-66.466)^2}{196&plus;171}=83.186$

$\therefore combined\;\;standard\;\;deviation=\sqrt{83.186}=9.121$