Question

Ages of Moviegoers Based on the data from Cumulative Review Exercise 7, assume that ages of moviegoers are normally distributed with a mean of 35 years and a standard deviation of 20 years.

a. What is the percentage of moviegoers who are younger than 30 years of age?

b. Find\({P_{25}}\), which is the 25th percentile.

c. Find the probability that a simple random sample of 25 moviegoers has a mean age that is less than 30 years.

d. Find the probability that for a simple random sample of 25 moviegoers, each of the moviegoers is younger than 30 years of age. For a particular movie and showtime, why might it not be unusual to have 25 moviegoers all under the age of 30?

Short answer

a.The percentage of moviegoers who are younger than 30 years of age is equal to 40.13%.

b.The 25^th percentile is equal to 21.6 years.

c. The probability of the mean age to be less than 30 years is equal to 0.1056.

d. The probability of all the 25 moviegoers being less than 30 years of age is approximately equal to 0. The audience for a specific film and showing time is not a random sample. Some films and showtimes are popular among children and teenagers.

Step-by-step solution

Given information

The ages of moviegoers are normally distributed with a mean equal to 35 years and a standard deviation equal to 20 years.

Percentage

a.

Let X be the age of the moviegoers.

It is given that\(X \sim N\left( {\mu = 35,\sigma = 20} \right)\)

The proportion of moviegoers who are younger than 30 years of age is given by:

\(\begin{aligned} P\left( {X < 30} \right) &= P\left( {\frac{{X - \mu }}{\sigma } < \frac{{30 - \mu }}{\sigma }} \right)\\ &= P\left( {\frac{{X - 35}}{{20}} < \frac{{30 - 35}}{{20}}} \right)\\ &= P\left( {Z < - 0.25} \right)\\ &= 0.4013\end{aligned}\)

Thus, the percentage of moviegoers younger than 30 years of age is equal to 40.13%.

Percentile

b.

To compute the 25^th percentile of X, the z score corresponding to the p-value of 0.2 looked from the standard normal table and is equal to:

\(\begin{aligned} P\left( {Z < z} \right) &= 0.25\\z &= - 0.67\end{aligned}\)

The corresponding value of the age for z-score equal to -0.67 is computed below:

\(\begin{aligned} z &= \frac{{x - \mu }}{\sigma }\\ - 0.67 &= \frac{{x - 35}}{{20}}\\x &= 21.6\end{aligned}\)

Thus, the 25^th percentile is equal to 21.6 years.

Distribution of sample mean

c.

The mean of the sample of ages is normally distributed with mean\(\left( \mu \right)\)and standard deviation\(\left( {\frac{\sigma }{{\sqrt n }}} \right)\).

Here, n=25.

The probability of the mean age to be less than 30 years is computed below:

\(\begin{aligned} P\left( {\bar X < 30} \right) &= P\left( {\frac{{\bar X - \mu }}{{\frac{\sigma }{{\sqrt n }}}} < \frac{{30 - \mu }}{{\frac{\sigma }{{\sqrt n }}}}} \right)\\ &= P\left( {z < \frac{{30 - 35}}{{\frac{{20}}{{\sqrt {25} }}}}} \right)\\ &= P\left( {z < - 1.25} \right)\\ &= 0.1056\end{aligned}\)

The probability of the mean age to be less than 30 years is equal to 0.1056.

Probability

d.

The probability of a moviegoer being less than 30 years of age is equal to 0.4013.

Let the sample size be 25.

The probability of all the 25 moviegoers to be less than 30 years of age is computed below:

\({\left( {0.4013} \right)^{25}} \approx 0.000\)

Thus, the probability of all the 25 moviegoers being less than 30 years of age is approximately equal to 0.

The audience for a specific film and showing time is not a random sample. Some films and showtimes are popular among children and teenagers.