Question

There are 48 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of 5 min and a standard deviation of 4 min. (Round your answers to four decimal places.)

(a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?(b) What is the (approximate) probability that the sample mean hardness for a random sample of 39 pins is at least 51?

Answer 1

a) 64.06% probability that he is through grading before the 11:00 P.M. TV news begins.

b) The hardness distribution is not given. But you would have to find s when n = 39, then the probability would be 1 subtracted by the pvalue of Z when X = 51.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (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 = (\sigma)/(√(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 n trials, the mean is
`\mu*n` and the standard deviation is
`s = \sigma√(n)`

In this question:

`n = 48, \mu = 48*5 = 240, s = 4√(48) = 27.71`

These values are in minutes.

(a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?

From 6:50 PM to 11 PM there are 4 hours and 10 minutes, so 4*60 + 10 = 250 minutes. This probability is the pvalue of Z when X = 250. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (250 - 240)/(27.71)`

`Z = 0.36`

`Z = 0.36` has a pvalue of 0.6406

64.06% probability that he is through grading before the 11:00 P.M. TV news begins.

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 39 pins is at least 51?

The hardness distribution is not given. But you would have to find s when n = 39(using the standard deviation of the population divided by the square root of 39, since it is not a sum here), then the probability would be 1 subtracted by the pvalue of Z when X = 51.