Let Y_1,...,Y_n be i.i.d. normal random variables with mean u and standard deviation o.
(a) Find the density function for Y(bar)=(Y_1+Y_2+...+Y_n)/2
(b) If o=4 and n=25, find P(|Y(bar)-u|>=1).
(c) If o=4, find the smallest n so that P(|Y(bar)-u|>=1)<=.01.
Solution
(a) I guess by Y(bar) you wanted the mean of the n random variables, i.e. (Y_1+Y_2+...+Y_n)/n
instead of(Y_1+Y_2+...+Y_n)/2
We know that if n random variables are iid normal, then any linear combination also would be a
normal distribution. So we need to calculate the mean and variance to get the density.
Mean = (u+u+..+u ntimes)/n =u by linearity of expectation,
variance = (o^2+o^2+..+o^2 n times)/n^2 = o^2/n since the covariances are zero (iid random
vars)
density is given by exp[-(x-mean)^2/2*variance]/sqrt(2*pi*variance) for all real x
(b) [Y(bar) - u]/sqrt(o^2/n) follows standard normal distribution.
P(|Y(bar)-u|>=1) =P(|Y(bar)-u|/(4/5)>=5/4) Now look into two sided tail probability for 5/4 in
the distribution table for standard normal for the answer.
(c) Same as above but now we have n unknown but we know the two sided tail probability. Look
for the approximate n based on the two sided tail probability .01.
Let Y_1,...,Y_n be i.i.d. normal random variables with mean u and st.pdf
1. Let Y_1,...,Y_n be i.i.d. normal random variables with mean u and standard deviation o.
(a) Find the density function for Y(bar)=(Y_1+Y_2+...+Y_n)/2
(b) If o=4 and n=25, find P(|Y(bar)-u|>=1).
(c) If o=4, find the smallest n so that P(|Y(bar)-u|>=1)<=.01.
Solution
(a) I guess by Y(bar) you wanted the mean of the n random variables, i.e. (Y_1+Y_2+...+Y_n)/n
instead of(Y_1+Y_2+...+Y_n)/2
We know that if n random variables are iid normal, then any linear combination also would be a
normal distribution. So we need to calculate the mean and variance to get the density.
Mean = (u+u+..+u ntimes)/n =u by linearity of expectation,
variance = (o^2+o^2+..+o^2 n times)/n^2 = o^2/n since the covariances are zero (iid random
vars)
density is given by exp[-(x-mean)^2/2*variance]/sqrt(2*pi*variance) for all real x
(b) [Y(bar) - u]/sqrt(o^2/n) follows standard normal distribution.
P(|Y(bar)-u|>=1) =P(|Y(bar)-u|/(4/5)>=5/4) Now look into two sided tail probability for 5/4 in
the distribution table for standard normal for the answer.
(c) Same as above but now we have n unknown but we know the two sided tail probability. Look
for the approximate n based on the two sided tail probability .01
![Let Y_1,...,Y_n be i.i.d. normal random variables with mean u and standard deviation o.
(a) Find the density function for Y(bar)=(Y_1+Y_2+...+Y_n)/2
(b) If o=4 and n=25, find P(|Y(bar)-u|>=1).
(c) If o=4, find the smallest n so that P(|Y(bar)-u|>=1)<=.01.
Solution
(a) I guess by Y(bar) you wanted the mean of the n random variables, i.e. (Y_1+Y_2+...+Y_n)/n
instead of(Y_1+Y_2+...+Y_n)/2
We know that if n random variables are iid normal, then any linear combination also would be a
normal distribution. So we need to calculate the mean and variance to get the density.
Mean = (u+u+..+u ntimes)/n =u by linearity of expectation,
variance = (o^2+o^2+..+o^2 n times)/n^2 = o^2/n since the covariances are zero (iid random
vars)
density is given by exp[-(x-mean)^2/2*variance]/sqrt(2*pi*variance) for all real x
(b) [Y(bar) - u]/sqrt(o^2/n) follows standard normal distribution.
P(|Y(bar)-u|>=1) =P(|Y(bar)-u|/(4/5)>=5/4) Now look into two sided tail probability for 5/4 in
the distribution table for standard normal for the answer.
(c) Same as above but now we have n unknown but we know the two sided tail probability. Look
for the approximate n based on the two sided tail probability .01](https://image.slidesharecdn.com/lety1-230324034835-35e4195a/85/Let-Y_1-Y_n-be-i-i-d-normal-random-variables-with-mean-u-and-st-pdf-1-320.jpg)
