Question

Your friend Kate is never on time and is regularly anywhere from 3 to 21 minutes late. Let X
represent the length of time in minutes that Kate is late and assume it has a uniform distribution.a) Draw the probability density of X.
b) Find the probability that Kate will be no more than 10 minutes late.
c) Find the probability that Kate will be between 15 and 20 minutes late.
d) Find the probability that Kate will be exactly 12 minutes late.
e) You and Kate are going to an 8:00 PM movie so she agreed to pick you up from your house at 7:25 PM. If it takes 21 minutes to drive from your house to the movie theatres and 7 minutes to buy tickets and snacks, what is the probability that you will be seated before the movie begins?

Answer 1

The definition of uniform distribution can be found on the website, for example. The figure shows the image and the definition of constant distribution taken from site https://mathworld.wolfram.com/UniformDistribution.html .

Answers:

a) In our case $a=3; b=21 min$ , respectively the probability density function $P(X)=\frac{1}{18}$ ;

b) The probability that Kate will be no more than 10 minutes late $P_b$ is the area under the curve $P(X)$ from the value $X=a$ to the set value of $X=10 min$.Thus, $P_b=\frac{(10-3)}{18}=\frac{7}{18}$ ;

c) Similarly $P_c=\frac{20-15}{18}=\frac{5}{18}$ ;

d) The probability that Kate will be exactly $X=12 min$ late equals $P_d=P(X)=0$ .

e) If everything goes without delays, you and Katya will be seated in the cinema at time T=7:25+0:21+0:07=7:53. Before the start of the session, you could have 7 minutes, but Katya is usually at least 3 minutes late, so your margin is 4 minutes, and the probability of being in time is $P_e=\frac{4}{18}=\frac{2}{9}$ .