Question

The length of time a full length movie runs from opening to credits is normally distributed with a mean of 2.2 hours and standard deviation of 0.2 hours. Calculate the following:A random movie is between 1.8 and 2.0 hours.A movie is shorter than 1.6 hours.

A movie is longer than 2.4 hours.

Answer 1

The distribution of a movie length is dfined by $\frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{(L-L_0)^2}{2 \sigma^2}}$ , where L is a movie length, $L_0 = 2.2$ is a mean movie length, and $\sigma = 0.2$ is a standard deviation. By substitution $x = \frac{L - L_0}{\sigma}$ we can convert this distribution to the normal standard distribution $\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}}$ .

Then the probability for a random movie to be between 1.8 and 2.0 hours is $P(1.8 < L < 2.0) = \int_{1.8}^{2.0}{\frac{1}{\sqrt{2 \pi} 0.2} e^{-\frac{(L-2.2)^2}{2*0.2^2}} dL} = \int_{-2}^{-1}{\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} dx} = \int_{1}^{2}{\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} dx}$

The last integral is equals $\frac{1}{2}(erf(\frac{2}{\sqrt{2}}) - erf(\frac{1}{\sqrt{2}})) =\frac{1}{2}(0.9544997361-0.68268949213) = 0.13590512198$

The probabbility for a movei to be shotter than 1.6 hour $P(L<1.6) = \int_{0}^{1.6}{\frac{1}{\sqrt{2 \pi} 0.2} e^{-\frac{(L-2.2)^2}{2*0.2^2}} dL} = \int_{-11}^{-3}{\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} dx}= -\int_{3}^{11}{\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} dx}$

and $P(L<1.6) = (erf(\frac{11}{\sqrt{2}}) - erf(\frac{3}{\sqrt2}))/2 = (1-0.99730020393)/2 =0.00134989803$

The probability for a movie to be longer than 2.4 hours:

$P(L>2.4) = \int_{2.4}^{\infin}{\frac{1}{\sqrt{2 \pi} 0.2} e^{-\frac{(L-2.2)^2}{2*0.2^2}} dL} == \int_{1}^{\infin}{\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} dx} = (erf(\infin) - erf(\frac{1}{\sqrt2}))/2 = (1-0.68268949213)/2 = 0.15865525393$