The distribution of the number of viewers for the American Idol television broad

Question

The distribution of the number of viewers for the American Idol television broadcasts follows the normal distribution with a mean of 30 million and a standard deviation of 10 million. What is the probability that next week's show will: Have between 34 and 43 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Have at least 26 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Exceed 56 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Explanation / Answer

THE MEAN = 30(MILLION)

HERE THE STANDARD DEVIATION = 10(MILLION)

IN ALL THE PARTS WE NEED TO FIND THE PROBABILITY BY CALCULATING THE Z SCORE

Z SCORE = (X-MEAN)/STANDARD DEVIATION

A)

For x = 34 , z = (34 - 30) / 10 = 0.4 and for x = 43, z = (43 - 30) / 10 = 1.3

Hence P(34 < x < 43) = P(0.4 < z < 1.3) = [area to the left of z = 1.3] - [area to the left of 0.4]
NOW WE WILL TAKE THE VALUE FROM THE Z TABLE
= 0.9032 - 0.6554 = 0.2478

B) AT LEAST 26

THAT MEANS WE NEED TO FIND P(X>=26) =

For x = 26, z = (26 - 30) / 10 = -0.4

Hence P(x > 26) = P(z > -0.4) = [total area] - [area to the left of -0.4]

1 - [area to the left of -0.4]

now from the z table we will take the value of z score = -0.4

1 - 0.3446 = 0.6554

C) EXCEED 56

THEREFORE WE NEED TO FIND P(X>56) =

For x = 56, z = (56 - 30) / 10 = 2.6

Hence P(x > 56) = P(z > 2.6) = [total area] - [area to the left of 2.6]

1 - [area to the left of 2.6]

now from the z table we will take the value of z score = 2.6

1 - 0.9953 = 0.0047

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