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

Answer:

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

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:

$\mu = 4.5, \sigma = 1.2$

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

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

$Z = \frac{6 - 4.5}{1.2}$

$Z = 1.25$

$Z = 1.25$ has a pvalue of 0.8944

X = 4

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

$Z = \frac{4 - 4.5}{1.2}$

$Z = -0.42$

$Z = -0.42$ 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