Question

1. What is the mean and standard deviation for the number of calls between 3:10PM and 3:15PM from the above problem?

1b. The probability that Alice arrives on time is 80%. The probability that Bob arrives on time is 40%. If neither affects the other, what is the probability that both Alice and Bob arrive on time? What is the probability they do not arrive on time?

1c. In a survey of middle school students, 70% play basketball and 40% play basketball and football. Of those that play basketball, what percentage also plays football?

Answer 1

For question number 1, the full problem is not given. If you could provide the full problem, I would solve it for you.

Question (1b)

The probability that Alice arrives on time is 80%. The probability that Bob arrives on time is 40%.

Let us denote

A=Event that Alice arrives on time

B=Event that Bob arrives on time.

So, it is given that

Now, Alice's arrival and Bob's arrival are independent of each other.

Question a)

Probability that both Alice and Bob arrive on time.

=P(A and B)

=P(A)P(B)

=0.8*0.4

=0.32

Question b)

Probability that they both do not arrive on time

=P(A' and B')

=P(A')P(B')

=(1-P(A))*(1-P(B))

=(1-0.8)*(1-0.4)

=0.2*0.6

=0.12

Question (1c)

In a survey of middle school students, 70% play basketball and 40% play basketball and football.

Let us denote

F=Event that a student plays football

B=Event that a student plays basketball.

So, given that

P(B)=0.7

P(F and B)=0.4

Now, we have to find the percentage of students who play football, of those who play basketball.

So, basically what we have to find is

P(F|B)

By the definition of conditional probability, it becomes

=P(F and B)/P(B)

=0.4/0.7

=0.5714

So, the corresponding percentage is approximately 0.5714*100, ie. 57.14%.

The percentage of students who play football, of the students who play basketball is 57.14%.