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 23 million with a standard deviation of 6 million.

Have between 27 and 35 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Have at least 15 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Exceed 37 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Explanation / Answer

here mu=23,sigma=6

answer of a)

for 27 the z-value, z1=(x-mu)/sigma=(27-23)/6=0.75

for 35 the z=value, z2=(35-23)/6=2

p(z1<z<z2)=p(z2)-p(z1)=0.9772-0.7733=0.2039

answer part (b)

for 15 , the z-value z=(15-23)/6=-1.5

p(z<-2)=0.0668

answer of part(c)

for 37, z-value z=(37-23)/6=2.33

p(z>2.33)=1-p(z<2.33)=1-.9901=.0099

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