Let Y1 Yn be a random sample of size n from a normal dis

Let Y1, . . . , Yn be a random sample of size n from a normal distribution with mean

Solution

Basically, you have a normal distribution with some mean and some standard deviation. You need to find the z value that will encompass 95% of the samples.

What do you know???

you know the variance is 10; so the s.d. is sqrt(10)
you know the z value associated with a (two tailed) distibution containing 95%; 1.96
you (should) know how to compute a z score
(x - mu) / s.d.
you (should) know how to compute the s.d. of the mean (s,d,/sqrt(n))

you do not know the mean
you need to compute the number (n) required

the formula is

(x - mu)/s.d. = 1,96

(x - mu)/s.d/sqrt(n) (where s.d./sqrt(n) is the standardard deviation OF THE MEAN, not the population

x - mu should be .5 (you don\'t NEED to know the sample mean or population mean--just the difference between them

s.d. is sqrt(10) = 3.16

.5 / 3.16/sqrt(n) = 1.96