Question

Americans receive an average of 22 Christmas cards each year. Suppose the number of Christmas cards is normally distributed with a standard deviation of 6. Let X be the number of Christmas cards received by a randomly selected American. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(,)

b. If an American is randomly chosen, find the probability that this American will receive no more than 25 Christmas cards this year.

c. If an American is randomly chosen, find the probability that this American will receive between 24 and 27 Christmas cards this year.

d. 76% of all Americans receive at most how many Christmas cards? (Please enter a whole number)

Answer 1

Solution :

Given that ,

mean = = 22

standard deviation = = 6

a.

X N (22 , 6)

b.

P(x 25)

= P[(x - ) / (25 - 22) / 6]

= P(z 0.5)

= 0.6915

Probability = 0.6915

c.

P(24 < x < 27) = P[(24 - 22)/ 6) < (x - ) /  < (27 - 22) / 6) ]

= P(0.33 < z < 0.83)

= P(z < 0.83) - P(z < 0.33)

= 0.7967 - 0.6293

= 0.1674

Probability = 0.1674

d.

Using standard normal table ,

P(Z z) = 76%

P(Z 0.71) = 0.76

z = 0.71

Using z-score formula,

x = z * +

x = 0.71 * 6 + 22 = 26

Christmas cards = 26