Question

The average time a subscriber spends reading the local newspaper is 49 minutes. Assume the standard deviation is 16 minutes and that the times are normally distributed. For the 10% who spend the most time reading the paper, how much time do they spend?

Answer 1

Answer:

They spend at least 69.48 minutes reading the paper.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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

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

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.

In this problem, we have that:

$\mu = 49, \sigma = 16$

For the 10% who spend the most time reading the paper, how much time do they spend?

They spend at least X minutes, in which X is the value of X when Z has a pvalue of 0.90. So it is X when Z = 1.28.

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

$1.28 = \frac{X - 49}{16}$

$X - 49 = 16*1.28$

$X = 69.48$

They spend at least 69.48 minutes reading the paper.