Suppose that X is the time it takes a randomly chosen professor in a university department to type and send a standard letter of recommendation for a past student who is applying for a graduate degree program. Suppose X has a normal distribution, and assume the mean is 10.5 minutes and the standard deviation 3 minutes. If you were to take a random sample of 50 professors and measure the time it takes each one of them to type and send such a standard letter of recommendation, what is the probability that their mean time is less than 9.5 minutes?

Respuesta :

Answer:

0.0091 = 0.91% probability that their mean time is less than 9.5 minutes.

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 normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 p-value, 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

The mean is 10.5 minutes and the standard deviation 3 minutes.

This means that [tex]\mu = 10.5, \sigma = 3[/tex]

Sample of 50:

This means that [tex]n = 50, s = \frac{3}{\sqrt{50}}[/tex]

What is the probability that their mean time is less than 9.5 minutes?

This is the pvalue of Z when X = 9.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{9.5 - 10.5}{\frac{3}{\sqrt{50}}}[/tex]

[tex]Z = -2.36[/tex]

[tex]Z = -2.36[/tex] has a pvalue of 0.0091

0.0091 = 0.91% probability that their mean time is less than 9.5 minutes.