Question

The graph shows the distribution of the amount of time (in minutes) people spend watching TV shows on a popular streaming service. The distribution is approximately Normal, with a mean of 71 minutes and a standard deviation of 15 minutes.
A graph titled Streaming T V has time (minutes) on the x-axis, going from 11 to 131 in increments of 15. The highest point of the curve is at 71.
What percentage of people spend between 41 and 56 minutes watching TV shows on this streaming service?
a- 13.5%
b- 34%
c- 47.5%
d- 95%
answer:a 13.5

Answer 1

Using the normal distribution, it is found that the percentage of people spend between 41 and 56 minutes watching TV shows on this streaming service is:

a) 13.5%

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

In this problem:

• The mean is of 71, hence $\mu = 71$

• The standard deviation is of 15, hence $\sigma = 15$

The proportion between 41 and 56 minutes is the p-value of Z when X = 56 subtracted by the p-value of Z when X = 41, hence:

X = 56:

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

$Z = \frac{56 - 71}{15}$

$Z = -1$

$Z = -1$ has a p-value of 0.16

X = 41:

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

$Z = \frac{41 - 71}{15}$

$Z = -2$

$Z = -2$ has a p-value of 0.025

0.16 - 0.025 = 0.135

0.135 x 100% = 13.5%

Hence option a is correct.

A similar problem is given at https://brainly.com/question/25769446