Question

Americans receive an average of 20 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( 20 , 6 ) b. If an American is randomly chosen, find the probability that this American will receive no more than 24 Christmas cards this year. 0.7486 c. If an American is randomly chosen, find the probability that this American will receive between 21 and 26 Christmas cards this year. d. 66% of all Americans receive at most how many Christmas cards? (Please enter a whole number)

Answer 1

The distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.

Probability

a. Distribution

X ~ N (20 , 6)

b. P(x ≤24)

= P[(x - μ ) / σ  (24 - 20) / 6]

= P(z  ≤0.67)

=  0.74857

=0.7486

Hence:

Probability = 0.7486

c. P(21 < x < 26)

= P[(21 - 26)/ 6) < (x - μ  ) / σ   < (24 - 20) / 6) ]

= P(-0.83 < z < 0.67)

= P(z < 0.67) - P(z < -0.)

=  0.74857- 0.2033

= 0.54527

Hence:

Probability =0.54527

d. Using standard normal table ,

P(Z < z) = 66%

P(Z < 0.50) = 0.66

z = 0.50

Using z-score formula,

x = z×  σ +  μ

x = 0.50 × 6 + 20 = 23

23 Christmas cards

Therefore the distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.

Learn more about probability here:brainly.com/question/24756209

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