# Answer on Statistics and Probability Question for Angela

Question #16552

normal-shaped distribution has μ = 80 and σ = 15.

a. sketch the distribution of sample means for samples of n = 25 scores from this

population

b. what are the z-score values that form the boundaries for the middle 95% of the distribution of the sample means?

c. compute the z-scores for M = 89 for a sample of n = 25 scores. Is this sample mean in the middle 95% of the distribution? Is this sample mean in the middle 99% of the distribution?

d. compute the z-score for M = 84 for a sample of n = 25 scores. Is this sample mean in the middle 95% of the distribution?

a. sketch the distribution of sample means for samples of n = 25 scores from this

population

b. what are the z-score values that form the boundaries for the middle 95% of the distribution of the sample means?

c. compute the z-scores for M = 89 for a sample of n = 25 scores. Is this sample mean in the middle 95% of the distribution? Is this sample mean in the middle 99% of the distribution?

d. compute the z-score for M = 84 for a sample of n = 25 scores. Is this sample mean in the middle 95% of the distribution?

Expert's answer

(a) the distributionof sample means for samples of n = 25 scores from this population will be the

normal distribution with mean of 80 and standard deviation of 15/sqrt(25) = 3.

(b) p (z(- x) < z< z( x)) = 0.95 = 2*p(0 < z < z(x)) => z(x) = 1.96 => z = -1.96;

1.96

(c) z(M) = (89 -80)/(15/sqrt(25)) = 3 > 1.96 => this sample mean is not in the middle 95%

of the distribution. For the 99% confidence interval z = -2.575, 2.575. z(M) =

3 does not belong to this interval. So, this sample mean is not in the middle

99% of the distribution.

(d) z(M) = (84 -80)/(15/sqrt(25)) = 4/3 < 1.96 => this sample mean is in the middle 95%

of the distribution.

normal distribution with mean of 80 and standard deviation of 15/sqrt(25) = 3.

(b) p (z(- x) < z< z( x)) = 0.95 = 2*p(0 < z < z(x)) => z(x) = 1.96 => z = -1.96;

1.96

(c) z(M) = (89 -80)/(15/sqrt(25)) = 3 > 1.96 => this sample mean is not in the middle 95%

of the distribution. For the 99% confidence interval z = -2.575, 2.575. z(M) =

3 does not belong to this interval. So, this sample mean is not in the middle

99% of the distribution.

(d) z(M) = (84 -80)/(15/sqrt(25)) = 4/3 < 1.96 => this sample mean is in the middle 95%

of the distribution.

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