Question

A fifth grade teacher gives homework every night in both mathematics and language arts. The time to complete the mathematics homework has a mean of 30 minutes and a standard deviation of 3 minutes. The time to complete the language arts homework has a mean of 40 minutes and a standard deviation of 4 minutes. Assuming both distributions of homework time are Normally distributed, what is the probability that the average time it takes to complete both homework assignments is greater than 82 minutes?

Answer 1

0.82% probability that the average time it takes to complete both homework assignments is greater than 82 minutes

Explanation:

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.

Math homework:

`\mu_M = 30, \sigma_M = 3`

Arts homework:

`\mu_A = 40, \sigma_A = 4`

What is the probability that the average time it takes to complete both homework assignments is greater than 82 minutes?

Here, we have a sum of normal variables. The mean will be the sum of the means, while the standard deviation is the square root of the sum of the variances. So

`\mu = \mu_M + \mu_A = 30 + 40 = 70`

`\sigma = √(\sigma_M^2 + \sigma_A^2) = √(3^2+4^2) = 5`

This probability is 1 subtracted by the pvalue of Z when X = 82. So

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

`Z = (82 - 70)/(5)`

`Z = 2.4`

`Z = 2.4` has a pvalue of 0.9918

1 - 0.9918 = 0.0082

0.82% probability that the average time it takes to complete both homework assignments is greater than 82 minutes