A random sample of size 28 is to be selected from a population that has a mean =
ID: 3396570 • Letter: A
Question
A random sample of size 28 is to be selected from a population that has a mean = 51 and a standard deviation of 8.
(a) This sample of 28 has a mean value of x, which belongs to a sampling distribution. Find the shape of this sampling distribution.
(b) Find the mean of this sampling distribution. (Give your answer correct to nearest whole number.)
(c) Find the standard error of this sampling distribution. (Give your answer correct to two decimal places.)
(d) What is the probability that this sample mean will be between 40 and 54? (Give your answer correct to four decimal places.)
(e) What is the probability that the sample mean will have a value greater than 45? (Give your answer correct to four decimal places.)
(f) What is the probability that the sample mean will be within 2 units of the mean? (Give your answer correct to four decimal places.)
Explanation / Answer
Given:
A random sample of size 28 is to be selected from a population that has a mean = 51 and a standard deviation of 8.
a)This sample of 28 has a mean value of x, which belongs to a sampling distribution. Find the shape of this sampling distribution.
The shape of the distribution is normal.
b)
The mean of the sampling distribution = E(sample mean) = Population mean =51
c) Standard error of the sampling distribution = standard deviation/sqrt(n) = 8/sqrt(28) = 1.51
d)
The probability that this sample mean will be between 40 and 54
=P(40 <SM<54) = P[(40-51)/1.51 < z< (54-51)/1.51}=P(-7.285< z < 1.987) = 0.9765
( where SM=sample mean)
e) Probability that the sample mean will have a value greater than 45 is
P(SM>45) = P{z > (45-51)/1.51}=1- P{z <=-3.9735} =0.9999 =1 (approximately)
f) probability that the sample mean will be within 2 units of the mean
2 units of the mean = (51-2, 51+2) = ((49, 53)
Required probability is =P(49 <SM<53) = P[(49-51)/1.51 < z< (53-51)/1.51}=P(-1.325< z < 1.325) = 0.8147
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