Question

A study was conducted on the minuets people spend browsing websites. The time spent is normally distributed with a mean of 4.5 minutes and a standard deviation of 1.2 about what percentage of people spent between 4 and 6 minutes browsing the sites?

Answer 1

55.72% of people spent between 4 and 6 minutes browsing the sites

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 = (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 = 4.5, \sigma = 1.2`

What percentage of people spent between 4 and 6 minutes browsing the sites?

This is the pvalue of Z when X = 6 subtracted by the pvalue of Z when X = 4. So

X = 6

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

`Z = (6 - 4.5)/(1.2)`

`Z = 1.25`

`Z = 1.25` has a pvalue of 0.8944

X = 4

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

`Z = (4 - 4.5)/(1.2)`

`Z = -0.42`

`Z = -0.42` has a pvalue of 0.3372

0.8944 - 0.3372 = 0.5572

55.72% of people spent between 4 and 6 minutes browsing the sites