Probability and Probability Distributions Problems

Q.1. Among male birds of a species, 20% have a particular gene. Among females of the species, 10% have the
gene. The males comprise 40% of all the birds of the species (thus, females comprise 60%).

p.1.a. What is the probability a randomly selected bird of this species has the gene?
p.1.b. What is the probability the bird is male, given it has the gene?

Given: P ( G | M ) = 0.20 P ( G | F ) = 0.10 P ( M ) = 0.40 P ( M ) = 0.60

p.1.a. P ( G ) = P ( GM ) + P ( GF ) = P ( M ) P ( G | M ) + P ( F ) P ( G | F )
= 0.40(0.20) + 0.60(0.10) = 0.08 + 0.06 = 0.14
P ( GM ) 0.08
p.1.b. P ( M | G ) = = = .5714
P (G ) 0.14

Q.2. You are going to a foreign nation to conduct your research. On a weather website you see that the average
high temperature during the period you will be there has been historically 20 degrees Celsius. What is the
average in degrees Fahrenheit? Hint: F=32+(9/5)C

9 9 9

F = 32 +   C C = 20   F = 32 +   C = 32 +   20 = 32 + 36 = 68
5 5   5  

Q.3. In a population of 100 watt light bulbs manufactured by a company, 80% (0.80 as a proportion) have
lifetimes exceeding 800 hours. An inspector samples 10 bulbs at random. What is the probability that all 10
bulbs’ lifetimes exceed 800 hours?

Y  # out of 10 lightbulbs sampled with lifetimes > 800 hours

10 
Y ~ Bin ( n = 10,  = 0.8 ) P (Y = 10 ) =   ( 0.8 ) (1 − 0.8 ) = ( 0.8 ) = 0.1074
10 10 −10 10

10 
Q.4. Among students taking a standardized exam, scores are normally distributed with a mean of 550 and
standard deviation 100. What proportion of the students score above 700?

 Y −  700 − 550 
Y ~ N (  = 550,  = 100 ) P (Y  700 ) = P  Z =  = 1.50  = .0668
  100 
Q.5. A sample of 5 animals of a particular species is selected at random from the population being managed in a
wildlife refuge. If 5% (0.05) of the population have a particular trait, what is the probability that none of the 5
tested have the trait?

5
Y ~ Bin ( n = 5,  = 0.05 )  P (Y = 0 ) =   (.05 ) (1 − .05 ) = .7738
0 5− 0

0
Q.6. Based on the following contingency table, complete the following parts:

Concussion No Concussion Total

Male 40 25596 25636
Female 60 27107 27167
Total 100 52703 52803

p.6.a. Among Males, what is Probability of concussion?

p.6.b. Among Females, what is Probability of concussion?

40 60
P (C | M ) = = .00156 P ( C | F ) = = .00221
25636 27167

Q.7. A quality engineer in a factory is interested in the proportion of all computer chips that her assembly line
produces that meet a particular quality requirement. She selects a sample of 100 chips and finds that 85 pass the
test. This means that =0.85 is the proportion of all chips the assembly line produces that meet the requirement.
FALSE

Q.8. A sample of 6 animals of a particular species is selected at random from the population being managed in a
wildlife refuge. If 15% (0.15) of the population have a particular trait, what is the probability that none of the 6
tested have the trait?

.3771

Q.9. A new simpler (LA) test for bird flu is compared with the existing gold standard (HI) test, with the
following results. Assume the gold standard (HI) test is completely accurate:

HI (+) HI (-) Total

LA (+) 664 2 666
LA (-) 84 80 164
Total 748 82 830

p.9.a. What is the probability a person with the bird flu tests positive on the LA test?
p.9.b. What is the probability a person without the bird flu tests negative on the LA test?

P(LA+ | HI+) = 664/748 = .8877 P(LA- | HI-) = 80/82 = .9756

Q.10. A magazine publisher includes winning coupons for an advertised product in 1% ( = 0.01) of the October issues of
the magazine. A national firm buys a random sample of n=500 of the issues (assume the total number of issues is many
times > 500). Let Y be the number of winning coupons the firm receives. Give the expected value of Y, and the
probability that they get 1 or fewer winning coupons.

p.10.a. E(Y) = n = 500(.01) = 5 p.10.b. P(Y ≤ 1) = .99500 + 500(.011)(.99499) = .00657 + .03318 = .03975

Q.11. Two reviewers (Rev1 and Rev2) are compared by their positive and negative reviews of 1000 movies:

Rev2 (+) Rev2(-) Total

Rev1 (+) 400 200 600
Rev1 (-) 100 300 400
Total 500 500 1000
p.11.a. What is the probability both reviewers give the same review? (400 + 300) / 1000 = .70

p.11.b. What is the probability Reviewer 2 was positive, given Reviewer 1 was negative? 100 / 400 = .25

Q.12. Heights of adult males (cm) are approximately normally distributed with M = 167 and M = 6. Heights of adult
females (cm) are approximately normally distributed with F = 160 and F = 5.

p.12.a. What proportion of males are taller than 175 cm?

p.12.b. What is the 95th %-ile among female heights?

YM ~ N ( M = 167,  M = 6 ) YF ~ N (  F = 160,  F = 5 )
p.11.a. P (YM  175) = P ( Z M  1.33) = .0918
p.11.b. P ( Z  1.645) = .05 = P (YF  160 + 1.645(5) = 168.225)

Q.13. In a population of people on a “Singles Cruise”, 60% are females and 40% are males. Among the females, 20% are
actually married (and cheating on their spouse), among males, 40% are married.

p.13.a. What is the probability a randomly selected “single” is actually married? .60(.20) + .40(.40) = .28

p.13.b. Given the randomly selected “single” is married, what is the probability that it is male? .16 / .28 = .5714

Q.14 The probability of randomly selecting the correct response on a multiple choice question with five choices is 0.20
(assuming zero knowledge). Suppose an exam consists of 6 multiple choice questions, each with five choices.

p.14.a. How many correct responses would you expect a student to pick by randomly selecting answers? 6(.2) = 1.2

p.14.b. What is the probability a student gets none of the questions correct? (1-0.2)6 = .2621

Q.15 A bridge holds up to 25 cars. It is known the weight of individual cars is normally distributed with  = 2100 lbs. and
 = 500 lbs.

p.15.a. What is the sampling distribution of the sample mean weight for n=25 cars?

p.15.b. If the maximum load for which the bridge is designed is 80,000 lbs., what is the probability a full load of cars
(n=25) will exceed the design limits?

  500 
p.15.a. Y ~ N (  = 2100,  = 500 ) Y ~ N  Y =  = 2100,  Y = = = 100 
 n 25 
 Y − Y 3200 − 2100 
( )
25
1 25 1
p.15.b.  Yi  80000 Y =  Yi = (80000) = 3200 P Y  3200 = P  Z =  = 11   0
25 i =1 25  Y 100 
i =1  

Q.16. A diagnostic test has 95% sensitivity (the probability a person with the condition tests positive = 0.95) and 95%
specificity (the probability a person without the condition tests negative = 0.95). In a population of people given the test,
1% of the people have the condition (probability a person has the condition = 0.01).
p.16.a. What proportion of the people will test positive?

p.16.b. Given a person has tested positive, what is the probability he/she has the condition?

Given: P ( C + ) = .01 P (T + | C + ) = .95 P (T − | C − ) = .95

 P ( C − ) = 1 − P ( C + ) = 1 − .01 = .99 P (T + | C − ) = 1 − P (T − | C − ) = 1 − .95 = .05
p.16.a. P (T + ) = .01(.95) + .99(.05) = .0095 + .0495 = .0590

p.16.b. P ( C + | T + ) =
.0095
= .1610
.0590

Q.17. In a population of birds on a desert island, 40% are Red, 30% are Yellow, 20% are Black, and 10% are Green.
Among the Red birds, 5% have a genetic trait, among the Yellows, 10% have the trait, among the Blacks, 20% have the
trait, and among the Greens, 25% have the trait. Let T be the event that a bird has the trait. Complete the following table.

Color P(Color) P(T|Color) P(T&Color) P(Color|T)

Red 0.4000 0.0500 0.0200 0.1739
Yellow 0.3000 0.1000 0.0300 0.2609
Black 0.2000 0.2000 0.0400 0.3478
Green 0.1000 0.2500 0.0250 0.2174
Total 1.0000 #N/A 0.1150 1.0000

Q.18. A soccer player on the UF soccer team has a probability of scoring on a penalty kick against a particular goalie of 
= 0.80. Suppose in the course of a game, she has 3 penalty kicks against the goalie.

p.18.a. What is the probability she scores on all 3 kicks (assuming independence)? (0.8)3 = .512

p.18.b. What is the probability she fails to score on all 3 kicks? (.2)3 = .008
^
p.18.c. Suppose each day she attempts 64 penalty kicks, and observes  , the sample proportion of successful attempts.
^
The sampling distribution of  is approximately:

Shape: Normal Mean Standard Error ___________________________

Shape: Normal Mean:  ^ =  = 0.8


^
Standard Error: SE  =  ^ =

 (1 −  )
n
=
(0.8)(0.2)
64
= 0.05
Q.19. In a large-scale study of SAR levels in cellphone models for 3 brands (LG, Motorola, Nokia), one
characteristic reported was whether the model had low SAR levels. There were 457 total models studied, with
P(LG) = 0.25, P(Motorola) = 0.40, and P(Nokia) = .35. Among LG phones, the proportion with low SAR levels
is 0.24, among Motorola, the proportion is 0.20, and among Nokia, the proportion is 0.28.

p.19.a. Compute the probability a randomly selected phone has a low SAR level.

p.19.b. Compute the probability of each brand, given the phone has a low SAR level.

LS  Low SAR P ( LS ) = .25(.24) + .40(.20) + .35(.28) = .060 + .080 + .098 = .238

.060 .080 .098
P ( LG | LS ) = = .2521 P ( M | LS ) = = .3361 P ( N | LS ) = = .4118
.238 .238 .238

Q.20. A movie preview is shown independently to n = 10 people. Each person is then offered the chance to
purchase a ticket. The probability any individual person, after seeing the preview, will buy the ticket is 0.4.

p.20.a. What is the probability that exactly 4 of the 10 buy the ticket? (10!/(4!6!))(0.4)4(0.6)6 = .2508

p.20.b. What is the mean and variance of the number of the 10 people to buy the ticket?

Mean: 10(0.4) = 4 Variance: 10(0.4)(0.6) 2.4

Q.21. Body Mass Indices (BMI) for National Hockey League (NHL) players are approximately normally distributed with
a mean of 26.50 and standard deviation of 1.45.

p.21.a. What is the probability a randomly selected NHL player has a BMI below 25.0?

p.21.b. Between what 2 BMI levels do the middle 95% of all NHL players fall?

p.21.c. What is the sampling distribution of sample means of sample size = 25 from this population? Give the distribution
symbolically and draw a graph of it.

Y ~ N ( 26.50,1.45 )
p.21.a. P (Y  25 ) = P ( Z  −1.03) = P ( Z  1.03) = .1515
p.21.b. P ( Z  1.96 ) = P ( Z  −1.96 ) = .025  P ( −1.96  Z  1.96 ) = .95
 P ( 23.658 = 26.50 − 1.96(1.45)  Y  26.50 + 1.96(1.45) = 29.342 ) = .95
 1.45 
p.21.c. n = 25  Y ~ N  26.50 , = 0.29  Y  2 Y  ( 25.92, 27.08 )
 25 

Q.22. During the years 1960-2015 the mean and standard deviation for July high temperatures at Orlando
International airport were F = 91.4°F and F = 8.3°F, respectively. You have friends arriving from a country
that uses the Celsius system, where °C = 0.56F – 17.78. Report to your friends the mean and standard deviation
in terms of degrees Celsius. Mean = 33.404 SD = 4.648
Q.23. Chicago food establishments are classified by 3 levels of Risk (High, Medium, and Low). The
proportions (probabilities) are: P(High) = .65, P(Medium) = .22, P(Low) = .13. The probabilities of Failing
inspection are .21 among High risk, .23 among Medium Risk, and .33 among Low Risk.
Note: weird numbers (I may have reversed labels). Answers are based on numbers as they are stated.

p.23.a. Compute the probability a randomly selected establishment Fails inspection. .2300

p.23.b. Compute the probability of each risk type, given the establishment fails inspection.

P(High Risk | Fail) = .5935 P(Medium Risk | Fail) = .2200 P(Low Risk | Fail) = .1865

Q.24. An examination is given with n = 5 multiple-choice questions, each with 4 choices, and 1 correct answer.
A student arrives for the exam completely un-prepared and will randomly guess on each question.

p.24.a. What is the probability the student will get at least 1 correct answer? 1-(1-.25)4 = 1-.3164 = .6836

p.24.b. What are the mean and standard deviation of the number correct answers if this exam was given many
times to people randomly guessing answers? Mean = 5(0.25) = 1.25 SD = (5(.25)(.75))1/2 = .9682

Q.25. Body Mass Indices (BMI) for English Premier League (EPL) football players are approximately normally
distributed with a mean of 23.00 and standard deviation of 1.70.

p.25.a. What is the probability a randomly selected EPL player has a BMI above 24.5? P(Z ≥ 0.88) = .1894

p.25.b. Between what 2 BMI levels do the middle 50% of all EPL players fall?

P(Z ≥ .675) ≈ .25 => 23.00 ± 0.675(1.70) ≡ 23.00 ± 1.15 ≡ (21.85,24.15)

p.25.c. What is the sampling distribution of sample means of sample size = 16 from this population? Give the distribution
symbolically and draw a graph of it.

 1.70 
Y ~ N  23.00, = 0.425  23.00  2(0.425)  23.00  0.85  ( 22.15, 23.85 )
 16 

Q.26. The June monthly rainfall totals (in inches) for a sample of 5 Orlando years were: 10, 5, 5, 6, 9.

Give the sample mean, median, and standard deviation of the monthly rainfall totals (show all work).

10 + 5 + 5 + 6 + 9 35
Mean: y = = = 7 Ordered: 5,5,6,9,10 Median = 6
5 5
(10 − 7 ) + ( 5 − 7 ) + ( 5 − 7 ) + ( 6 − 7 ) + ( 9 − 7 )
2 2 2 2 2
22
SD: s = = = 2.345
5 −1 4
Q.27. Elite female hammer thrower Anita Włodarczyk has a competitive mean distance thrown of 73.36 meters
and standard deviation of 2.74 meters. Translate her mean and standard deviation to feet. (1 foot = 0.3048
meters).

1 foot = 0.3048 meters => 1 meter = 1/0.3048 = 3.281 feet

Mean: 3.281(73.36) = 240.69 SD: 3.281(2.74) = 8.99