Random Variable and Green Pods Essay Example
Chapter - Section | Problem(s) | 5 – 2 | 10, 14 | 5 – 3 | 12, 32 | 5 – 4 | 18 |
5-2 #10,14
10. Eye Color Groups of five babies are randomly selected. In each group, the random variable x is the number of babies with green eyes (based on data from a study by Dr. Sorita Soni at Indiana University). (The symbol 0+ denotes a positive probability value that is very small). X | P(x) | 0 | .528 | 1 | .360 | 2 | .098 | 3 | .013 | 4 | .001 | 5 | 0+ |
P(x) = .528 + .360 + .098 + .013 + .001 + (0+) = 1
This is a probability distribution because P(x) = 1 for the data given
Standard Deviation = 4.9344
14. Range Rule of Thumb for Unusual Events: Use the range rule of thumb to identify a range of values containing the usual number of peas with green pods. Based on the result, is it unusual to get only one pea with a green pod? Explain.
Probabilities of Numbers of Peas with Green Pods among 8 offspring Peas. X(Number of Peas with Green Pods) | P(x) | 0 | 0+ | 1 | 0+ | 2 | .004 | 3 | .023 | 4 | .087 | 5 | .208 | 6 | .311 | 7 | .267 | 8 | .100 |
P(x) = (0+) + (0+) + .004 +.023 + .087 + .208 + .311 + .267 + .100 = 1
The above data showcases a probability distribution because P(x) is equal to 1
Standard Deviation: = 1.25
5-3 #12, 32
#12 The above data question shows that it is a Binomial Distribution because it is data taken from a population sample
#32 The probability that at least 1 of 12 people become delinquent is less than person therefore the bank should NOT prepare for delinquency amongst the borrowers.
5-4 #18 A. Mean = 3.21 a. Standard Deviation = 30.352 B. Based on the prior results it is not unusual to find that among 420095 people that there would be 135 cases of cancer because it is almost exact to the deviated amount. C. It suggests that though there is a risk of cancer, it is not that large of a mitigating factor that can cause widespread
5-2 #10,14
10. Eye Color Groups of five babies are randomly selected. In each group, the random variable x is the number of babies with green eyes (based on data from a study by Dr. Sorita Soni at Indiana University). (The symbol 0+ denotes a positive probability value that is very small). X | P(x) | 0 | .528 | 1 | .360 | 2 | .098 | 3 | .013 | 4 | .001 | 5 | 0+ |
P(x) = .528 + .360 + .098 + .013 + .001 + (0+) = 1
This is a probability distribution because P(x) = 1 for the data given
Standard Deviation = 4.9344
14. Range Rule of Thumb for Unusual Events: Use the range rule of thumb to identify a range of values containing the usual number of peas with green pods. Based on the result, is it unusual to get only one pea with a green pod? Explain.
Probabilities of Numbers of Peas with Green Pods among 8 offspring Peas. X(Number of Peas with Green Pods) | P(x) | 0 | 0+ | 1 | 0+ | 2 | .004 | 3 | .023 | 4 | .087 | 5 | .208 | 6 | .311 | 7 | .267 | 8 | .100 |
P(x) = (0+) + (0+) + .004 +.023 + .087 + .208 + .311 + .267 + .100 = 1
The above data showcases a probability distribution because P(x) is equal to 1
Standard Deviation: = 1.25
5-3 #12, 32
#12 The above data question shows that it is a Binomial Distribution because it is data taken from a population sample
#32 The probability that at least 1 of 12 people become delinquent is less than person therefore the bank should NOT prepare for delinquency amongst the borrowers.
5-4 #18 A. Mean = 3.21 a. Standard Deviation = 30.352 B. Based on the prior results it is not unusual to find that among 420095 people that there would be 135 cases of cancer because it is almost exact to the deviated amount. C. It suggests that though there is a risk of cancer, it is not that large of a mitigating factor that can cause widespread