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

Question

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 31 million with a standard deviation of 10 million.

What is the probability next week's show will:

a. Have between 35 and 42 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

b. Have at least 23 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

c. Exceed 50 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Explanation / Answer

Mean ( u ) =31
Standard Deviation ( sd )=10
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 35) = (35-31)/10
= 4/10 = 0.4
= P ( Z <0.4) From Standard Normal Table
= 0.65542
P(X < 42) = (42-31)/10
= 11/10 = 1.1
= P ( Z <1.1) From Standard Normal Table
= 0.86433
P(35 < X < 42) = 0.86433-0.65542 = 0.2089                  
b.
P(X < 23) = (23-31)/10
= -8/10= -0.8
= P ( Z <-0.8) From Standard Normal Table
= 0.2119                  
P(X > = 23) = (1 - P(X < 23)
= 1 - 0.2119 = 0.7881                  
c.
P(X > 50) = (50-31)/10
= 19/10 = 1.9
= P ( Z >1.9) From Standard Normal Table
= 0.0287                  

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