Suppose that the random sample is taken from a normal distribution N(8,9), and the random sample is between 1 to 25.
Find the distribution of the sample mean.
Find probability that the sample mean is less than or equal to 8.8 and the sample variance is less than or equal to 12.45, where the probabilities are independent.
Find probability that the sample mean is less than 8+(.5829)S, where S is the sample standard deviation.
Let n = 25
Then as per the central limit theorem:
Sampling distribution will be normally distributed.
with
mean, = 8
=
= 3/5
= 0.6
P(8.8) = P(Z(8.8-8)/0.6)
= P(Z1.333)
= 0.9087
We know that
P(<(24*12.45/9) = P(<33.2)
= 0.9
P(<8+(.5829)s) =P((-8)/s)<0.5829)
= P((-8)/(s/5))<0.5829*5)
= P(Z<2.9145)
= 0.9982
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