Q15. A variable assumes the values 1,2,....,7 with frequencies $\small 1^2,2^2,...,7^2$ respectively. Calculate the mean of the variable.
Ans:[Hint: $\small \bar x= \frac{\sum_{i=1}^{7} x_{i} f_{i}}{\sum_{i=1}^{7} f_{i}}$,=$\small\frac{ (1^3+2^3+...+7^3)}{1^2+2^2+....+7^2}$.]
Q16.Find the mean of$\small 7,77,777,7777,...... $up to $\small p th$ term.
Answer:
We need to find $\small 7+77+777+7777+...$ p th term
Let, $\bar x$ is the mean
$\bar x$= $\small \frac{(7+77+777+7777+...)}{p}$
$\bar x$=$\small {u}{p}$
let, $\small u$=$\small 7+77+777+7777+...$ p th term
or, $\small\frac{9}{7} u$=$\small 9+99+999+9999...$ p th term
or, $\small\frac{9}{7} u$=$\small (10-1)+(100-1)+(1000-1)+(10000-1)+...$ p th term
or, $\small\frac{9}{7} u$=$\small (10+100+1000+...)+...-(p)$ p th term
or, $\small\frac{9}{7} u$=$\small (10+10^2+10^3+...)+...-(p)$ p th term
or, $\small\frac{9}{7} u$=$\small \frac{10(10^p-1)}{9}-(p)$
Now, $\small\frac{u}{p}$
=$\small \frac{70(10^p-1)}{81}+ \frac{7}{9}$
Q17.Let the arithmetic mean of $\small x_{1},x_{2},...,x_{5}$ $\small(x_{1}\leq x_{2} \leq .... \leq x_{5})$ be equal to 10. Difine $\small y_{1},......,y_{10}$ as
$\small y_{1}=14, y_{2}=x_{1}$,
$\small y_{3}=y_{4}=x_{2}$,
$\small y_{5}=y_{6}=x_{3}$,
$\small y_{7}=y_{8}=x_{4}$,
$\small y_{9}=y_{10}=x_{5}$.
Show that the arithmetic mean of y can not be smaller than 10.4.
[Hint:$\small \sum_{i=1}^{5} x_{i}$= $\small 50$
$\small x_{1}\leq x_{2}$
$\small x_{1}\leq x_{3}$
$\small x_{1}\leq x_{4}$
$\small x_{1}\leq x_{5}$
Summing up both the side:
or, $\small 4 x_{1}\leq (x_{2}+x_{3}+x_{4}+x_{5})$
or, $\small 5 x_{1}\leq (x_{1}+x_{2}+x_{3}+x_{4}+x_{5})$
or, $\small 5 x_{1}\leq 50 $
or, $\small x_{1} \leq 10$
Q18. Suppose a variable assumes the values 0,1,2,....,n with frequencies proportional to the binomial coefficients $\small \binom{n}{0},\binom{n}{1},....,\binom{n}{n}$. Find the mean.
Ans: [Hint: $\small \sum_{i=1}^{n} x_{i}f_{i}$= $\small 0 \binom{n}{0} +..........+ n\binom{n}{n}$=$\small n2^{n-1})$, $\sum_{i=1}^{n}f_{i}$= $\small \binom{n}{0} +..........+ \binom{n}{n}$=$\small 2^{n})$]
Q19.The mean of 50 observations is found to be 18. on further verification it was detected that the values 22 and 16 wrongly taken as 19 and 24. Find the Corrected mean.
Ans: [Hint: $\small \sum_{i=1}^{n} x_{i}$=$\small 900$. Corrected $\small \sum_{i=1}^{n} x_{i}$=$\small 900+22+16-19-24 = 895$. Now find the corrected $\small \bar x$=$\small \frac{895}{50}$]
Q20. If The average monthly income of the workers in factory is Rs. 520. and the average monthly income of male workers and female workers are Rs. 540, Rs. 460. Respectively. Find the percentage of male and female workers.
Ans: [Hint:Let, Find the ratio,$\small n_{1}(male):n_{2}(female)$ from the equation $\small \frac{n_{1}540+n_{2}460} {n_{1}+n_{2}}=520$,
After that find the percentage of $\small n_{1},n_{2}$ ]
Q21.If the number of observations of two groups $\small G_{1}$ and $\small G_{2}$ are in the ratio 2:1. their arithmetic means are 8 and 128. Then write down the correct answer for A.M of the combined group among the alternatives.
i. 88
ii. 45$\small \frac{1}{3}$
iii.48
iv. None of the above
Ans: let the $\small G_{1}$ has $\small n_{1}$ number of observations and the $\small G_{2}$ has $\small n_{2}$ number of observations.
$\small n_{1}:n_{2}$=$\small 2:1$. So, $\small n_{1}=2 n_{2}$. Now Put this result into this equation of combined mean: $\small \bar x$=.$\small \frac{n_{1}\bar x_{1}+n_{2}\bar x_{2}} {n_{1}+n_{2}}$.
Tags: Giri Banarjee Statistics Solution, Central tendency question answer, Mean Median Mode