“Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad length follows th

Question

“Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad length follows the normal distribution with a population standard deviation of 2 seconds. Suppose we select a sample of 16 ads produced by Majesty.

a. What can we say about the shape of the distribution of the sample mean time?

b. What is the standard error of the mean time?

c. What percent of the sample means will be greater than 31.25 seconds?

d. What percent of the sample means will be greater than 28.25 seconds?

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?”

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Rylee 5 days 2021-09-15T22:57:35+00:00 1 Answer 0

Answers ( )

  1. Answer:

    a) \bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

    b) Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

    c) P(\bar X >31.25)=0.006=0.6\%

    d) P(\bar X >28.25)=0.9997=99.97\%

    e) P(28.25 <\bar X <31.25)=0.9936=99.36%

    Step-by-step explanation:

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.  

    The central limit theorem states that “if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large”.

    Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

    X\sim N(\mu =30,\sigma =2)

    We take a sample of n=16 . That represent the sample size.

    a. What can we say about the shape of the distribution of the sample mean time?

    From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

    \bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

    \bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

    b. What is the standard error of the mean time?

    The standard error is given by this formula:

    Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

    c. What percent of the sample means will be greater than 31.25 seconds?

    In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

    z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

    And we want to find this probability:

    P(\bar X >31.25)=1-P(\bar X<31.25)=1-P(Z<\frac{31.25-30}{0.5})=1-P(Z<2.5)=1-0.994=0.006=0.6\%

    d. What percent of the sample means will be greater than 28.25 seconds?

    In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

    z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

    And we want to find this probability:

    P(\bar X >28.25)=1-P(\bar X<28.25)=1-P(Z<\frac{28.25-30}{0.5})=1-P(Z<-3.5)=1-0.00023=0.9997=99.97\%

    e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?”

    We want this probability:

    P(28.25 <\bar X <31.25)=P(\frac{28.25-30}{0.5}<Z<\frac{31.25-30}{0.5})=P(-3.5<Z<2.5)=P(Z<2.5)-P(Z<-3.5)=0.9938-0.000233=0.9936=99.36\%

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