Answer to Question #118666 in Statistics and Probability for Michael

Question

The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standard deviation of 0.3 hours. Calculate the following :
a. A random movie is between 1.8 and 2.0 hours
b. A movie is longer than 2.3 hours
c. The length of movie that is shorter than 94% of the movies

Answer 1

Let X = the random variable denoting the length of time a full movie runs from opening to credits

We are given X ~ N($\mu=1.9,\sigma^2=0.3^2)$

Then we have,

Z = $\frac{X-\mu}{\sigma}$ ~ N(0, 1), Z is the standard normal variate

a. The probability that a random movie is between 1.8 and 2.0 hours

= P(1.8 $\leq$ X $\leq$ 2.0)

= P($\frac{1.8-1.9}{0.3}\leq\frac{X-1.9}{0.3}\leq\frac{2.0-1.9}{0.3})$

= P(- 0.33 $\leq$ Z $\leq$ 0.33)

= $\Phi(0.33)-\Phi(-0.33)$

= 0.6293 - 0.3707

= 0.2586

Answer: The probability that a random movie is between 1.8 and 2.0 hours is 0.2586.

b. The probability that a random movie is longer than 2.3 hours

= P(X > 2.3)

= P($\frac{X-1.9}{0.3}$ > $\frac{2.3-1.9}{0.3}$)

= P(Z > 1.33)

= 1 - P(Z $\leq$ 1.33)

= 1 - $\Phi$(1.33)

= 1 - 0.9082

= 0.0918

Answer: The probability that a random movie is longer than 2.3 hours is 0.0918.

c. Let the length of movie that is shorter than 94% of the movies be t.

Then P(X $\geq$ t) = 94%

i.e. P($\frac{X-1.9}{0.3}\geq\frac{t-1.9}{0.3}$) = 0.94

i.e. P(Z $\geq\frac{t-1.9}{0.3}$) = 0.94

i.e. 1 - P(Z < $\frac{t-1.9}{0.3}$) = 0.94

i.e. P(Z < $\frac{t-1.9}{0.3}$) = 1 - 0.94 = 0.06

i.e. $\Phi(\frac{t-1.9}{0.3})$ = 0.06 = $\Phi($-1.56)

i.e. $\frac{t-1.9}{0.3}$ = -1.56

i.e. t = 1.432 = 1.4 (rounded to 1 decimal place)

Answer: The length of movie that is shorter than 94% of the movies is 1.4 hours.