Question #325550

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Let's see how well you understood our discussion. At this point, I want you to



solve the following problems. Show your complete solution by following the step-by-



step procedure.



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?




1
Expert's answer
2022-04-10T15:32:34-0400

1.

a. (X>670)=1P(X670)=1P(Z67066035)=1P(Z67066035)1P(Z0.2857)=10.61409=0.38591(X>670)=1−P(X≤670) =1-P(Z\leq\dfrac{670-660}{35})=1−P(Z≤ \frac{ 670−660}{35} ​ ) \approx1-P(Z\leq0.2857)=1-0.61409=0.38591

b.(X>670)=1P(X670)=1P(Z67066035/10)=1P(Z67066035/10)1P(Z0.9035)0.1831(X>670)=1−P(X≤670) =1-P(Z\leq\dfrac{670-660}{35/\sqrt{10}})=1−P(Z≤ \frac{ 670−660}{35/\sqrt{10}} ​ ) \approx1-P(Z\leq0.9035)\approx0.1831

2.

P(X<68)=P(Z<68705.1/50)=P(Z<2.77)=0.0028P(X<68)=P(Z<\frac{68-70}{5.1/\sqrt{50}})=P(Z<-2.77)=0.0028





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