a)

To verify: If the normal model can be used to describe the number of Type AB donors among 320 donors.

Yes, the normal model can be used to describe the number of Type AB donors among 320 donors.

Given:

The percentage of people having Type AB blood=4% .

The total number of donations collected n=320 .

The number of Type AB donations among 320 donations x=26 .

Concept used:

A probability distribution where each trial has two possible outcomes known as a success and failure, each trial is independent and the probability of each trial is the same, then to calculate the probability of a certain number of successes occurring, a binomial probability distribution is used. A binomial model can be approximated to be a normal model if it satisfies the following conditions

np≥10, n(1-p)≥10 . where n is the total number of trials and p is the probability of success.

Calculation:

The total number of donations n=320 .

The probability that people have type AB blood p=0.04 .

The conditions to assume normal model are satisfied as shown below

np=320×0.04=12.8>10n(1−p)=320×(1−0.04)=307.2>10

Conclusion:

The conditions are satisfied, thus the number of Type AB donors can be assumed to be normally distributed.

b)

To calculate: The mean and standard deviation of the number of Type AB donors.

The mean and standard deviation of the number of donors is 12.8 and 3.5054 respectively.

Concept used:

The mean and standard deviation of the binomial model is calculated as shown below

mean=npstandard deviation=np(1−p)

Calculation:

The total number of donations n=320 .

The probability that people have type AB blood p=0.04 .

The mean and standard deviation are calculated as shown below

mean=np=320×0.04=12.8standard deviation=np(1−p)=320×0.04(1−0.04)=3.5054

Conclusion:

The mean and standard deviation of the distribution of the number of Type AB donors is 12.8 and 3.5054 respectively.

c)

To calculate: Whether the outcome of 26 Type AB donors is statistically significant.

The outcome of 26 type AB donors is statistically significant.

Concept used:

The Z score, the number of standard deviations away from the mean is calculated using the formula

Z=X−μσμ=meanσ=standard deviation

High values of the Z score are usually considered less likely to occur and statistically significant.Calculation:

The total number of donations n=320 .

The probability that people have type AB blood p=0.04 .

The number of people with Type AB blood x=26

The Z score is calculated as shown below

Z=X−μσZ=26−12.83.5054Z=3.77

Conclusion:

The Z score of 3.77 is very high and unusual. Thus the outcome of 26 type AB donors is statistically significant.

Chapter 10.4, Problem 39E

Chapter 10.4, Problem 41E

Chapter 10 Solutions

PRECALCULUS:GRAPHICAL,...-NASTA ED.

Ch. 10.1 - Prob. 1QR Ch. 10.1 - Prob. 2QR Ch. 10.1 - Prob. 3QR Ch. 10.1 - Prob. 4QR Ch. 10.1 - Prob. 5QR Ch. 10.1 - Prob. 6QR Ch. 10.1 - Prob. 7QR Ch. 10.1 - Prob. 8QR Ch. 10.1 - Prob. 9QR Ch. 10.1 - Prob. 10QR

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.1 - Prob. 31E Ch. 10.1 - Prob. 32E Ch. 10.1 - Prob. 33E Ch. 10.1 - Prob. 34E Ch. 10.1 - Prob. 35E Ch. 10.1 - Prob. 36E Ch. 10.1 - Prob. 37E Ch. 10.1 - Prob. 38E Ch. 10.1 - Prob. 39E Ch. 10.1 - Prob. 40E Ch. 10.1 - Prob. 41E Ch. 10.1 - Prob. 42E Ch. 10.1 - Prob. 43E Ch. 10.1 - Prob. 44E Ch. 10.1 - Prob. 45E Ch. 10.1 - Prob. 46E Ch. 10.1 - Prob. 47E Ch. 10.1 - Prob. 48E Ch. 10.1 - Prob. 49E Ch. 10.1 - Prob. 50E Ch. 10.1 - Prob. 51E Ch. 10.1 - Prob. 52E Ch. 10.1 - Prob. 53E Ch. 10.1 - Prob. 54E Ch. 10.1 - Prob. 55E Ch. 10.1 - Prob. 56E Ch. 10.1 - Prob. 57E Ch. 10.1 - Prob. 58E Ch. 10.1 - Prob. 59E Ch. 10.1 - Prob. 60E Ch. 10.1 - Prob. 61E Ch. 10.1 - Prob. 62E Ch. 10.2 - Prob. 1QR Ch. 10.2 - Prob. 2QR Ch. 10.2 - Prob. 3QR Ch. 10.2 - Prob. 4QR Ch. 10.2 - Prob. 5QR Ch. 10.2 - Prob. 6QR Ch. 10.2 - Prob. 7QR Ch. 10.2 - Prob. 8QR Ch. 10.2 - Prob. 9QR Ch. 10.2 - Prob. 10QR Ch. 10.2 - Prob. 1E Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - Prob. 5E Ch. 10.2 - Prob. 6E Ch. 10.2 - Prob. 7E Ch. 10.2 - Prob. 8E Ch. 10.2 - Prob. 9E Ch. 10.2 - Prob. 10E Ch. 10.2 - Prob. 11E Ch. 10.2 - Prob. 12E Ch. 10.2 - Prob. 13E Ch. 10.2 - Prob. 14E Ch. 10.2 - Prob. 15E Ch. 10.2 - Prob. 16E Ch. 10.2 - Prob. 17E Ch. 10.2 - Prob. 18E Ch. 10.2 - Prob. 19E Ch. 10.2 - Prob. 20E Ch. 10.2 - Prob. 21E Ch. 10.2 - Prob. 22E Ch. 10.2 - Prob. 23E Ch. 10.2 - Prob. 24E Ch. 10.2 - 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