There are 50 students in a calculus class. The amount of time needed for the instructor to grade a randomly chosen midterm exam paper is a random variable with a mean of 6 minutes and a standard deviation of 4 minutes. If grading times are independent, what is the probability that the instructor can finish grading in 4 and a half hours (round off to second decimal place)

Respuesta :

Answer:

14.46% probability that the instructor can finish grading in 4 and a half hours

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 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 [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.

For sums of n values from a distribution, the mean is [tex]n\mu[/tex] and the standard deviation is [tex]s = \sqrt{n}\sigma[/tex]

In this problem, we have that:

[tex]\mu = 6*50 = 300, s = \sqrt{50}*4 = 28.2843[/tex]

What is the probability that the instructor can finish grading in 4 and a half hours

Four and half hours is 4.5*60 = 270.

So this probability is the pvalue of Z when X = 270.

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

By the Central Limit Theorem

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

[tex]Z = \frac{270 - 300}{28.2843}[/tex]

[tex]Z = -1.06[/tex]

[tex]Z = -1.06[/tex] has a pvalue of 0.1446

14.46% probability that the instructor can finish grading in 4 and a half hours