Using the normal distribution and the central limit theorem, it is found that:
a) Since the distribution is skewed and the sample size is less than 30, the probability cannot be calculated.
b) There is a 0.1469 = 14.69% probability that the mean play time is more than 260 seconds.
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 mean and the standard deviation are given by, respectively:
[tex]\mu = 255, \sigma = 30[/tex]
In item a, according to the Central Limit Theorem, since the distribution is skewed and the sample size is less than 30, the probability cannot be calculated.
For item b, we have that n = 40 > 15, hence the standard error is given by:
[tex]s = \frac{30}{\sqrt{40}} = 4.74[/tex]
The probability is one subtracted by the p-value of Z when X = 260, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{260 - 255}{4.74}[/tex]
Z = 1.05
Z = 1.05 has a p-value of 0.8531.
1 - 0.8531 = 0.1469.
There is a 0.1469 = 14.69% probability that the mean play time is more than 260 seconds.
More can be learned about the normal distribution and the central limit theorem at https://brainly.com/question/24663213
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