Question

True or False: The amount of time it takes to complete an examination has a left skewed distribution with a mean of 65 minutes and a standard deviation of 8 minutes. If 64 students were randomly sampled, the probability that the sample mean of the sampled students exceeds 71 minutes is approximately 0.

Answer 1

Answer:

True

Step-by-step explanation:

To solve this question, 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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 65, \sigma = 8, n = 64, s = \frac{8}{\sqrt{64}} = 1$

If 64 students were randomly sampled, the probability that the sample mean of the sampled students exceeds 71 minutes is approximately 0.

This probability is 1 subtracted by the pvalue of Z when X = 71. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{71 - 65}{1}$

$Z = 6$

$Z = 6$ has a pvalue very close to 1.

1 - 1 = 0.

So approximately 0 probability that the sample mean of the sampled students exceeds 71 minutes.

The answer is true.