Answer to Question #196105 in Statistics and Probability for Betty

Question

Netflix is investing heavily in releasing their own productions under the “Netflix Originals” label. A call was sent out for scripts to be produced as Netflix Originals. The proposed runtimes of the scripts submitted were found to follow a normal distribution with mean 95 minutes and standard deviation 18 minutes. Netflix executives decided to keep the runtimes of their originals between 80 minutes and 125 minutes.

a) What is the probability that a randomly selected script has a runtime longer than 112 minutes?

b) Determine the 90th percentile value for the proposed runtimes of the scripts.

c) What is the probability that a randomly selected script will not satisfy the runtime requirements?

d) Records show that movies with runtimes in the top 13.45% of the accepted range have the most viewer retention, and make the most revenue. Therefore, what should the minimum runtime of a Netflix Original be if it is to maximise viewer retention?

Answer 1

Given that runtimes of the scripts follow a Normal distribution

with mean $\mu=95$ minutes and sd $\sigma$ = 18 minutes. Let us denote X as the runtime of a script.

a). The probability that a randomly selected script has a runtime longer than 112 minutes is P(X>112). We know that when X is normal $Z=\dfrac{X-\mu}{\sigma}$ follows a Standard Normal.

When $X=112, Z=\dfrac{112-95}{18}=0.9444$

P(X>112)=P(Z>0.9444)=1-P(Z<0.9444)

P(X>112)=1-0.8275 (Using the Excel function NORM.S.DIST(0.94444,1) to get this probability)

P(X>112)=0.1725

The probability that a randomly selected script has a runtime longer than 112 minutes is 0.1725

b). The 90th percentile value of the proposed runtimes of the scripts :

It is that point of the Z below which the area under the curve is 0.90

ie the point a such that P(X<a)=0.90

We get this value from the tables/Excel that is a=1.2816

ie when $Z=\dfrac{X-95}{18}=1.2816$

X=95+1.2816*18=118.0688

The 90th percentile value of the proposed runtimes of the scripts is 118.0688 minutes

c). The probability that a randomly selected script will not satisfy the rum time requirement :

The run time requirement is between 80 and 125 minutes. Therefore the probability is

$P(X<80)+P(X>125)$

When $X=80, Z=\dfrac{80-95}{18}=-0.8333$

P(X<80)=P(X<-0.8333)=0.2023

When $X=125, Z=\dfrac{125-95}{18}=1.6667$

P(X>125)=P(Z>1.6667)=1-P(Z<1.6667)

P(X>125)=1-0.9522=0.0478

The probability that a randomly selected script will not satisfy the rum time requirement=0.2023+0.0478=0.2501

The probability that a randomly selected script will not satisfy the rum time requirement is 0.2501

d). We know that the accepted range is 80-125 minutes.

From the previous calculations, the P(X<125)=0.9522.

The most viewer retention is the top 13.45 % of the accepted range and hence the minimum runtime to be on the top 13.45% of the accepted range is P(a<X<125)=0.1345

i.e. we get this point as P(X<125)-P(X<a)=0.1345

0.9522-P(X<a)=0.1345

i.e. P(X<a)=0.9522-0.1345=0.8177

For a standard Normal distribution, P(Z<0.9066)=0.8177

When $Z=0.9066, \dfrac{X-95}{18}=0.9066\Rightarrow X=95+16.3188=111.3188$

Therefore, the minimum runtime is 111.3188 minutes to be on the top 13.45% of the accepted range