Question

Carson's tablet has 2,000 videos. The play time for the videos is skewed to the right, with a mean of 135 seconds and a standard deviation of 25 seconds.

Part A: Can you accurately calculate the probability that the mean play time is more than 142 seconds for an SRS of 10 videos? Explain. (4 points)

Part B: If you take a random sample of 50 videos instead of 10, explain how the Central Limit Theorem allows you to find the probability that the mean play time is more than 142 seconds. Calculate this probability and show your work. (6 points)

Answer 1

Answer:

SEE BELOW

Step-by-step explanation:

A: Yes, it is 18.80%. Sample Mean= 135, Sample Standard deviation= 25/sq rt of 10 = 7.906, normalcdf(142, 999999, 135, 7.906) = .18796 × 100 = 18.796 = 18.80%

B: If we increase the sample size to 50, then, due to the Central Limit Therme, there should be less variation within the data (it should point to a more clear center). Sample Mean= 135, Sample Standard deviation= 25/sq rt of 50 = 3.535, normalcdf(142, 999999, 135, 3.535) = .02385 × 100 = 2.385 = 2.39%

Answer 2

The probability that the mean playtime is more than 142 seconds for an SRS of 10 videos is not accurately calculated and the probability that the mean playtime is more than 142 seconds is 0.0244.

Given :

• The total number of videos on Carson's tablet is 2000.
• The mean is 135 seconds.
• The standard deviation is 25 seconds.

A) The probability that the mean playtime is more than 142 seconds for an SRS of 10 videos is not accurately calculated because the sample size is too small due to which the central limit theorem cannot be applied.

B) According to the given data, the sample size is 50.

$P(\bar{X}>142)=P\left(\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n} }}>\dfrac{142-\mu}{\dfrac{\sigma}{\sqrt{n} }}\right)$

$P(\bar{X}>142)=P\left(Z>\dfrac{142-135}{\dfrac{25}{\sqrt{50} }}\right)$

$P(\bar{X}>142)=P\left(Z>1.9798\right)$

Using the z table the above expression becomes:

$P(\bar{X}>142)=0.0244$

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https://brainly.com/question/898534