Respuesta :
Using the normal distribution, it is found that about 81.8% of Internet users spend between 78 minutes and 174 minutes on social networking platforms each day.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
[tex]\mu = 142, \sigma = 32[/tex].
Th proportion of Internet users spend between 78 minutes and 174 minutes on social networking platforms each day is the p-value of Z when X = 174 subtracted by the p-value of Z when X = 78, hence:
X = 174:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{174 - 142}{32}[/tex]
Z = 1
Z = 1 has a p-value of 0.841.
X = 78:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{78 - 142}{32}[/tex]
Z = -2
Z = -2 has a p-value of 0.023.
0.841 - 0.023 = 0.818 = 81.8%.
81.8% of Internet users spend between 78 minutes and 174 minutes on social networking platforms each day.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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