Question 1159632: (Please help me. These are the last two questions for my homework. It's due in a few hours and I've been trying to solve them all day. Please be kind enough to help me.)
1. Suppose that a box contains 8 cameras and that 3 of them are defective. A sample of 2 cameras is selected at random. Define the random variable
X as the number of defective cameras in the sample
Write the probability distribution for
K
1.
2.
3.
P(X=k)
1.
2.
3.
What is the expected value for X?
2. Using all 1991 birth records in the computerized national birth certificate registry compiled by the National Center for Health Statistics (NCHS), statisticians Traci Clemons and Marcello Pagano found that the birth weights of babies in the United States are not symmetric (“Are babies normal?” The American Statistician, Nov 1999, 53:4). However, they also found that when infants born outside of the “typical” 37 - 43 weeks and infants born to mothers with a history of diabetes are excluded, the birth weights of the remaining infants do follow a Normal model with mean μ = 3432 g and standard deviation σ = 482 g. The following questions refer to infants born from 37 to 43 weeks whose mothers did not have a history of diabetes.
A. Compute the z-score of an infant who weighs 3779 g. (Round your answer to two decimal places.)
B. Approximately what fraction of infants would you expect to have birth weights between 2860 g and 4160 g? (Express your answer as a decimal, not a percent, and round to three decimal places.)
C. Approximately what fraction of infants would you expect to have birth weights above 4160 g? (Express your answer as a decimal, not a percent, and round to three decimal places.)
D.Approximately what fraction of infants would you expect to have birth weights above 4160 g? (Express your answer as a decimal, not a percent, and round to three decimal places.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 1 defective camera in the sample is either the first or the second
for the first the probability is (3/8) and second is (5/7) since it can't be defective
That combined probability is 2(15/56) since it can be the second that is defective alone. The probability is 30/56 or 15/28
Also the number of samples is 8C2=28
3C1*5C1 is the numerator
for 2 defective it is 3C2*5C0/8C2 or 3/28 probability. This is 3/8*2/7=6/56 =3/28
for 3 defective it is 0, since the sample is of 2.
for 0 defective it is (5/8)(4/7)=20/56=10/28, or the complement of the other probabilities
so for 0 prob 10/28=0 if multiplied for E(X)
for 1 prob is 15/28=15./28
for 2 prob is 3/28=6/28
Add them and E(x)=21/28 or 0.75 number of defective cameras in a sample of 2.
z=(x-mean)/sd so a would be (3779-3432)/482=0.72
for B the answer is 0.8169. Can also use calculator 2nd VARS 2 normalcdf (2860,4160,3432, 482) ENTER
for C the z is (4160-3432)/482=1.51. Probability z>1.51 is 0.0655 or 0.066.
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