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Sagot :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 71[/tex]
- The standard deviation is of 15, hence [tex]\sigma = 15[/tex]
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{56 - 71}{15}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a p-value of 0.16
X = 41:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{41 - 71}{15}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] 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
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