Question

"The time it takes for a statistics professor to mark a single midterm test is normally distributed with a mean of 4.8 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: 0.2743

Step-by-step explanation:

Given : The time it takes for a statistics professor to mark a single midterm test is normally distributed .

Mean : $\mu=4.8$

Standard deviation : $\sigma= 2.5$

Sample size : n=55

Let x be the random variable that represents the time it taken by statistics professor to mark a single midterm test .

The formula for z-score :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x= 5

$z=\dfrac{5-4.8}{\dfrac{2.5}{\sqrt{55}}}\approx0.6$

By using the standard normal distribution table ,the probability that he needs more than 5 hours to mark all of the midterm tests :-

$P(x>5)=P(z>0.6)=1-P(z\leq0.6)\\\\=1-0.7257469\approx0.2742531\approx0.2743$

Hence, the probability that he needs more than 5 hours to mark all of the midterm tests= 0.2743

Answer 2

Using the normal probability distribution and the central limit theorem, it is found that there is a 0.0262 = 2.62% probability that he needs more than 5 hours to mark all of the midterm tests.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

• By the Central Limit Theorem, for the sampling distribution of sample means of size n, the standard deviation is $s = \frac{\sigma}{\sqrt{n}}$

In this problem:

• Mean of 4.8 minutes, thus $\mu = 4.8$.
• Standard deviation of 2.5 minutes, thus $\sigma = 2.5$.
• Sample of 55, thus $n = 55, s = \frac{2.5}{\sqrt{55}}$.

More than 5 hours = 300 minutes is a sample mean above $\frac{300}{55} = 5.4545$, which means that the probability is 1 subtracted by the p-value of Z when X = 5.4545. Thus:

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

By the Central Limit Theorem

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

$Z = \frac{5.4545 - 4.8}{\frac{2.5}{\sqrt{55}}}$

$Z = 1.94$

$Z = 1.94$ has a p-value of 0.9738.

1 - 0.9738 = 0.0262

0.0262 = 2.62% probability that he needs more than 5 hours to mark all of the midterm tests.

A similar problem is given at https://brainly.com/question/15121634