The Emperor penguin, the most annoying of all penguins, has height (in inches) that follows a normal distribution with a mean of 40 and a standard deviation of 4.

If n = 25, would a sample mean of 42 be unusual?

(a) Yes because the z-score is 2.5
(b) Yes because the z-score is 0.5
(c) No because the z-score is 2.5
(d) No because the z-score is 0.5

(a) Yes because the z-score is 2.5

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Z-scores higher than 2 or lower than -2 are considered unusual

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

Would a sample mean of 42 be unusual?

We have to find Z when X = 42. So

By the Central Limit Theorem

A z-score of 2.5 means that a sample mean of 42 would be unusual.

So the correct answer is:

(a) Yes because the z-score is 2.5

(a) Yes because the z-score is 2.5

Step-by-step explanation:

z = (42 - 40)/(4/sqrt(25))

z = 2/0.8 = 2.5

|z| > 2 so unusual