Question

The time it takes for a statistics professor to mark a single midterm test is normally distributed with a mean of 5 minutes and a standard deviation of 2.5 minutes. There are 55 students in the professor's class. What is the probability that he needs more than 5 hours to mark all of the midterm tests

Answer 1

Answer:

8.85% probability that he needs more than 5 hours to mark all of the midterm tests

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sum of sizes n, the mean is $\mu*n$ and the standard deviation is $s = \sigma\sqrt{n}$

In this question:

$\mu = 55*5 = 275, \sigma = 2.5\sqrt{55} = 18.54$

What is the probability that he needs more than 5 hours to mark all of the midterm tests

5 hours is 5*60 = 300 minutes.

So this probability is 1 subtracted by the pvalue of Z when X = 300.

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

By the Central Limit Theorem

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

$Z = \frac{300 - 275}{18.54}$

$Z = 1.35$

$Z = 1.35$ has a pvalue of 0.9115

1 - 0.9115 = 0.0885

8.85% probability that he needs more than 5 hours to mark all of the midterm tests