Question #321032

1. The average number of milligrams (mg) of cholesterol in a cup of a certain brand of ice cream is 660 mg, the standard deviation is 35 mg. Assume the variable is normally distributed.


a. If a cup of ice cream is selected, what is the probability that the cholesterol content will be more than 670 mg?



b. If a sample of 10 cups of ice cream is selected, what is the probability that the mean of the sample will be larger than 670 mg?



2.In a study of the life expectancy of 400 people in a certain geographic region, the mean age at death was 70 years, and the standard deviation was 5.1 years. If a sample of 50 people from this region is selected, what is the probability that the mean life expectancy will be less than 68 years?

Expert's answer

1. We have a normal distribution, μ=600,σ=35.\mu=600, \sigma=35.


a.

Let's convert it to the standard normal distribution:

z=x−μσ=670−60035=2,z=\cfrac{x-\mu}{\sigma}=\cfrac{670-600}{35}=2,


P(X>670)=1−P(X<670)==1−P(Z<2)=P(X>670)=1-P(X<670)=\\ =1-P(Z<2)=

=1−0.9772=0.0228=1-0.9772=0.0228 ​(from z-table)


b.

Let's convert it to the standard normal distribution,

z=xˉ−μσ/n=670−60035/10=6.32.z=\cfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\cfrac{670-600}{35/\sqrt{10}}=6.32.

P(Xˉ>670)=1−P(Xˉ<670)==1−P(Z<6.32)=P(\bar{X}>670)=1-P(\bar{X}<670)=\\ =1-P(Z<6.32)=

=1−1=0=1-1=0 ​(from z-table)


2. We have a normal distribution, μ=70,σ=5.1,n=50.\mu=70, \sigma=5.1,n=50.

Let's convert it to the standard normal distribution,

z=xˉ−μσ/n=68−705.1/50=−2.77.z=\cfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\cfrac{68-70}{5.1/\sqrt{50}}=-2.77.

P(Xˉ<68)=P(Z<−2.77)=0.0028P(\bar{X}<68)=P(Z<-2.77)=0.0028 (from z-table)


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