The Probability That Both Have Type O Blood?

All human blood can be “ABO-typed” as one of O, A, B or AB, but the probability distribution of the types varies a bit among

groups of people. The table below gives the distribution of blood types for a randomly chosen person in the United States.

Blood Type / O / A / B / AB
U.S. Probability / 0.45 / 0.40 / 0.11 / ?

What is the probability that a randomly selected person in the US will have type AB blood?

Maria has type B blood. She can safely receive transfusions from people with blood types O and B. What is the probability that a randomly chosen American can donate blood to Maria?

Blood Type / O / A / B / AB
China Probability / 0.35 / 0.27 / 0.26 / 0.12

Choose a US citizen and a citizen of China at random, independently of one another. What is

the probability that both have type O blood?

Choose a US citizen and a citizen of China at random, independently of one another. What is

the probability that both have the same blood type?

(a) P(AB) = 1 - [P(O) + p(A) + P(B)] = 1 - (0.45 + 0.40 + 0.11) = 0.04
(b) The probability is P(O) + P(B) = 0.45 + 0.11 = 0.56
(c) The probability is 0.45 * 0.35 = 0.1575
(d) The probability is 0.45 * 0.35 + 0.40 * 0.27 + 0.11 * 0.26 + 0.04 * 0.12 = 0.2989

In each of the following situations, state whether or not the given assignment of probabilities to given outcomes is legitimate, that is, satisfies the rules of a probability distribution. (The assignment of probabilities need not follow common sense understanding of the outcomes!)

a. Roll a die and record the count of spots on the up face: P(1) = 0, P(2) = 1/6, P(3) = 1/3, P(4) = 1/3,

P(5) = 1/6, P(6) = 0.

b. Choose a college student at random and record sex and enrollment status: P(female-fulltime) = 0.56,

P(female-part time) = 0.24, P(male-fulltime) = 0.44, P(male- part time) = 0.17

c. Deal a card from a shuffled deck: P(Clubs) = 12/52, P(Diamonds) = 12/52, P(Hearts) = 12/52,

P(Spades) = 16/52.

(a) Not legitimate (b) Not legitimate (c) Not legitimate

3. Two psychologists surveyed 478 children in grades 4, 5 and 6 in elementary schools in Michigan. Among other

questions, they asked the students whether their primary goal was to get good grades, to be popular, or to be good

to be good at sports. One question of interest was whether boys and girls at this age had similar goals

Goals

Grades / Popular / Sports / Total
Boys / 117 / 50 / 60 / 227
Girls / 130 / 91 / 30 / 251
Total / 247 / 141 / 90 / 478

If a student is selected at random, what is the probability that the selected student is a boy?
If a student is selected at random, what is the probability that the selected student has as a goal to be good at sports?
Given that the selected student is a girl, what is the probability that she has a goal to be
popular?
Given that the selected student excels insports, what is the probability that the studentis a boy?
Find the probability that the selected student is a girlandhas a goal to be popular?
Find the probability that the selected passenger boyandhas a goal to begood at sports?
What is the probability that a student chosen at random has a goal to excel either at
academics or sports?
Are the events the selected student has a goal to be at the head of the class academically
and the selectedstudentis boy independent events? Explain.

(a) It is 227/478 = 0.4749
(b) It is 90/478 = 0.1883
(c) It is 91/251 = 0.3625
(d) It is 60/90 = 0.6667
(e) It is 91/478 = 0.1904
(f) It is 60/478 = 0.1255
(g) It is (247 + 90)/478 = 0.7050
(h) No, they are dependent events
Two events A and B are independent if P(A|B) = P(A)
Here, P(Grades |Boy) = 117/227 = 0.5154 and P(Grades) = 247/478 = 0.5167 which are different

Data from a National Health Survey show that men’s average weight is normally distributed with a mean of 170 pounds and a standard deviation of 30 pounds.

If a man is chosen at random, what is the probability that his weight is greater than 180 pounds?

If 100 men are randomly selected, what is the probability that their mean weight is greater than 180

pounds?

What is the probability that mean weight of the 100 men is less than 160

pounds?

(a) z = (x - μ)/σ = (180 - 170)/30 = 0.3333
P(x > 180 pounds) = P(z > 0.3333) = 0.3695
(b) z = (x-bar - μ)/(σ/√n) = (180 - 170)/(30/√100) = 3.3333
P(x-bar > 180 pounds) = P(z > 3.3333) = 0.0004
(c) z = (x-bar - μ)/(σ/√n) = (160 - 170)/(30/√100) = -3.3333
P(x-bar < 160 pounds) = P(z < -3.3333) = 0.0004

According to the College Board’s report, the average tuition and fees at four year private colleges and universities in the United States was $20,273 for the academic year 2002 – 2003 and the standard deviation was $4100. For a random sample of 100 four year private U.S. colleges:

a. What is the mean of the sampling distribution?

b. What is the standard deviation of the sampling distribution?

c. What is the probability that the mean tuition and fees of the sample is greater than $20,000?

d. What is the probability that the mean tuition and fees of the sample is less than $18,000?

e. What is the probability that the mean tuition and fees of the sample is within $410 of the population

mean?

(a) Mean of the sampling distribution = population mean = $20,273
(b) Standard deviation of the sampling distribution = σ/√n = 4100/√100 = $410
(c) z = (x-bar - μ)/(σ/√n) = (20000 - 20273)/410 = -0.6659
P(x-bar > $20000) = P(z > -0.6659) = 0.7473
(d) z = (x-bar - μ)/(σ/√n) = (18000 - 20273)/410 = -5.5439
P(x-bar < $18000) = P(z < -5.5439) = 0
(e) $410 is z = 1. So, the probability is P(-1 < z < 1) = 0.6827

The National Student Loan Survey collects data to examine questions related to the amount of money that borrowers owe. The survey selected a sample of 1280 borrowers who began repayment on their loans between 4 and 6 months prior to the survey. The mean debt for undergraduate study was $18,900 and the standard deviation was about $49,000.

Construct a 95% confidence interval for the true mean debt for the student borrowers.

n = 1280, x-bar = 18900, 49000, % = 95
Standard Error, SE = σ/n = 1369.59
z- score = 1.9600
Width of the confidence interval = z * SE = 2684.35
Lower Limit of the confidence interval = x-bar - width = 16215.65
Upper Limit of the confidence interval = x-bar + width = 21584.35
The confidence interval is [$16215.65, $21584.35]

Twenty-five volunteers who had developed cold symptoms within the last 24 hours were given zinc lozenges to take every 2 to 3 hours until their cold symptoms were gone. The mean overall duration of symptoms was 4.5 days with a standard deviation of 1.6 days.

The selected cases have a distribution which appears to be symmetric and bell-shaped.

Construct a 99% confidence interval for the mean overall duration of symptoms in a

population of individuals who used zinc lozenges.

In a previous test, the mean overall duration of symptoms for volunteers who were

told to drink an herbal tea for relief of their symptoms was 7 days. Based on your

confidence interval, do you think that using the lozenges reduces the overall duration

of symptoms compared to using the herbal tea?

(a) n = 25, x-bar = 4.5, 1.6, % = 99
Degrees of freedom = n - 1 = 24
Standard Error, SE = σ/n = 0.32
t- score = 2.7969
Width of the confidence interval = z * SE = 0.8950
Lower Limit of the confidence interval = x-bar - width = 3.6050
Upper Limit of the confidence interval = x-bar + width = 5.395
The confidence interval is [3.617 days, 5.40 days]
(b) Yes, using the lozenges reduces the overall duration of symptoms compared to using the herbal tea since 7 days lies outside the confidence interval computed above.

For the following problems:

1. State the Null Hypothesis and the Alternative Hypothesis

2. Determine the test statistic.

3. Determine the P-value

4. Make a decision regarding the hypotheses based on the P-value and the Level of

Significance.

Do middle-aged male executives have different average blood pressure than the general population? The NationalCenter for Health Statistics reports that the mean systolic pressure for males ages 35 to 44 is 128 and the standard deviation of this population is 15. The medical director of a company looks at the medical records of 72 company executives in this age group and finds that the mean systolic blood pressure in this sample is 126.07. Is this evidence that executive blood pressures differ from the national average? Use a 5% level of significance.

n = 72, μ = 128, x-bar = 126.07, σ = 15
(1) Ho: μ = 128 and Ha: μ ≠ 128
(2) z = (x-bar - μ)/(σ/√n) = -1.0918
(3) p- value = P(z > 1.0918) + P(z < -1.0918) = 0.2749
(4) Since 0.2749 > 0.05, we fail to reject Ho
There is no sufficient evidence that executive blood pressures differ from the national average

In fiscal 1996, the U.S. Agency for International Development provided 238,300 metric tons of corn soy blend (CSB) for development programs and emergency relief in countries throughout the world. CSB is a highly nutritious, low-cost fortified food that is partially cooked and can be incorporated into different food preparations by the recipients. The specifications for the CSB state that the mixture should contain 2 pounds of vitamin premix for every 200 pounds of product. These specifications are designed to produce a mean vitamin C content in the final product of 40mg per 100g. A sample of 8 readings from a production run has a mean of 22.50mg per 100g with a standard deviation of 7.19mg per gram. Is there significant evidence that this production run does not meet the specifications? Use a 1% level of significance.

n = 8, μ = 40, x-bar = 22.50, s = 7.19
(1) Ho: μ = 40 and Ha: μ ≠ 40
(2) t = (x-bar - μ)/(s/√n) = -6.884
(3) For 7 degrees of freedom, p- value = P(t > 6.884) + P(t < -6.884) = 0.0002
(4) Since 0.0002 < 0.01, we reject Ho and accept Ha
It appears that this production run does not meet the specifications

According to the National Institute for Occupational Safety and Health, job stress poses a

major threat to the health of workers. A national survey of restaurant employees found that

75% said that job stress had a negative impact on their personal lives. A sample of 100

employees of a restaurant chain finds that 68 answer “Yes” when asked: “Does job stress

have a negative impact on your personal life?” Is this good reason to think that the proportion

of all employees of this chain who would say “Yes” differs from the national proportion,

p0 = 0.75? Use a 5% level of significance.

n = 100, x = 68, p = 0.75
(1) Ho: p = 0.75 and Ha: p ≠ 0.75
(2) p’ = x/n = 0.68
SE = √[p(1 - p)/n] = 0.0433
z = (p’ - p)/SE = -1.6166
(3) p- value = P(z > 1.6166) + P(z < -1.6166) = 0.1060
(4) Since 0.1060 > 0.05, we fail to reject Ho
There is no sufficient reason to think that the proportion of all employees of this chain who would say “Yes” differs from the national proportion.