Question

The average time an individual reads online national news reports is 49 minutes. Assume the standard deviation is 16 minutes and that the times are normally distributed. What is the probability someone will spend no more than 30 minutes reading online national news reports?

Answer 1

Answer:

11.70% probability someone will spend no more than 30 minutes reading online national news reports.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 49, \sigma = 16$

What is the probability someone will spend no more than 30 minutes reading online national news reports?

This is at most 30 minutes, so 30 or less minutes. This is the pvalue of Z when X = 30.

So

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

$Z = \frac{30 - 49}{16}$

$Z = -1.19$

$Z = -1.19$ has a pvalue of 0.1170.

So there is a 11.70% probability someone will spend no more than 30 minutes reading online national news reports.