Find the Probability density function for the price of a ticket.

Question

If I have the following scenario:

• The age $X$ of a person going to the cinema is normally distributed with a mean = $22$ and varianz = $5$ going to a boxing match

• People under the age of 17 have to pay $10$ $

• From 17 to 18 they have to pay $12$ $

• And everyone above the age of 18 has to pay $15$ $

I want to calculate the PDF and mean for the price. My idea was to calculate the probability for a person within each of the three possible intervals. But I don't know how to and would appreciate any help regarding this problem!

Answer 1

Find probabilities under the normal curve for intervals $(0,17).[17,18),[18,100).$

Because the specified normal distribution has a bit of probability below $0,$ adjust the above probabilities so that they sum to $1,$ to obtain the PDF.

Multiply PDF by ticket prices and sum to get the expected value $\$14.05$of the distribution.

p = diff(pnorm(c(0,17,18, 100), 22, 5)); p
[1] 0.15864984 0.05320014 0.78814460
pdf = p/sum(pdf);  pdf
[1] 0.15865070 0.05320043 0.78814887
sum(pdf)
[1] 1
tkt = c(10,12,15)
mean = sum(tkt*pdf);  mean
[1] 14.04715

Addendum per comments showing how to use age distribution $A\sim\mathsf{Norm}(\mu=22,\sigma=4)$ to get probability of in interval $[17,18).$ $$P(17 \le A < 18) = P\left(\frac{17-22}{4} \le\frac{A-22}{4}< \frac{18-22}{4}\right) \\ =P(-1.25 \le Z < -1) = 0.1587 - 0.1056 = 0.0531,$$ where the numbers at the end come from printed normal tables. I don't know exactly the style of normal CDF table you may have, so I can't say exactly how to do it. But note that $0.0531$ here is very close to $0.05320014$---about as close as you're likely to get with printed normal tables.