Question

Ch 5 Statistics Binomial Probability/Poisson Probability For full credit, show your work step by step and circle your final answer. Then Upload in Blackboard (you can do just one time) (Round the final answer to 4 decimal digits)

2. The lengths of pregnancies are normally distributed with a mean of 273 days and a standard deviation of 20 days. If 50 women are randomly selected, find the probability that they have a mean pregnancy between 273 days and 275 days.

3. The body temperatures of adults are normally distributed with a mean of 98.6° F and a standard deviation of 0.60° F. If 25 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° F.

4. The average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22 pounds (Source: American Dietetic Association). If a sample of 60 individuals is randomly selected, find the probability that the mean of the sample will be Less than 200 pounds. I need help with 2, 3, & 4 please

Answer 1

2)

μ =273

σ = 20

n = 50

Z = X - μ /(σ /√n)

P(273 < X < 275)

= P[ 273 -273/(20/√50) < z < 275-273/(20/√50) ]

= P( 0 < z < 0.71)

= 0.2611 [standard normal distribution table]

3)

μ = 98.6

σ =0.6

n = 25

Z = X - μ/( σ /√n)

= 98.4 -98.6 /(0.6/√25)

= -1.67

P( X > 98.4)

= P(z > -1.67)

= 0.4525 + 0.5 [standard normal distribution table]

= 0.9525

4)

μ = 196

σ = 22

n=60

X=200

Z = X - μ /( σ/ √n)

= 200 -196 /(22/√60)

= 1.41

P(X < 200)

= P( Z < 1.41)

= 0.4207 + 0.5 [standard normal distribution table]

= 0.9207