2. Assume that a discrete random variable has the values of 5, 10, 15, 20, 25, a

Question

2. Assume that a discrete random variable has the values of 5, 10, 15, 20, 25, and 30. Furthermore, assume that the value 30 has probability of 0.50 associated with it, but each of the other values is equally likely.

A. What is the probability distribution?

B. What is the expected value of this variable?

C. What is the standard deviation of this variable?

D. What is the probability that this variable is 20 or more?

E. What is the probability that this variable is more than 20?

F. What is the probability that such a random variable has the value of 20 or less?

G. What is the probability that the variable has values between 18 and 28?

H. What is the interval of values that are within one standard deviation of the mean?

I. What is the probability that this variable has values within one standard deviation of the mean?

3. A small daycare center for young children has only 3, 4, 5, 6, or 7 children in attendance on any one day. On half of the days there are three children in attendance and on a quarter of the days there are 4 children in attendance. The probability on 5 and on 6 children in attendance is 0.10 each, with the remainder of the probability on the event that 7 children will attend.

A. What is the variable?

B. What are the values of the variable?

C. What is the probability distribution of the variable?

D. What is the expected number of children who will attend the small daycare center?

E. If an adult must be present for every 3 children, what is the probability that the daycare center will need 3 adults?

F. If an adult must be present for every 3 children, what is the probability that the daycare center will need 2 adults?

G. If an adult must be present for every 3 children, what is the probability that at least 2 adults are needed?

H. In an adult must be present for every 3 children, what is the probability that only one adult is needed?

Explanation / Answer

2.

a)

As the probabilities must sum up to 1, then the first 5 values of x must have p = 0.10. Hence,

[ANSWER]

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b)

Consider:

Thus,  
  
E(x) = Expected value = mean =    22.5 [ANSWER]

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c)

As

Var(x) = E(x^2) - E(x)^2 =    81.25

Then

s(x) = sqrt [Var(x)] =    9.013878189 [ANSWER]

**************

d)

P(20 or more) = P(20) + P(25) + P(30) = 0.1+0.1+0.5 = 0.7 [ANSWER]

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