STAT243 LS: Intro to Probability and Statistics
Final Exam, Mar 20, 2017
KEY
This is a 110-min exam. Students may use a page of note (front and back), and a
calculator, but nothing else is allowed.
1. A normal distribution is:
A) skewed right
B) skewed left
C) bi-modal
D) symmetric
2. Scientists discovered a new group of proteins in an animal species. They found that the
distribution of the number of amino acids these proteins were made of was approximately
normal with mean 530 and standard deviation 80.
About what percent of these new proteins will be between 370 and 690 amino acids long?
A) 68%
B) 95%
C) 99.7%
D) 32%
3. In a binomial setting, which of the following must NOT be true?
A) There must be a fixed number of observations.
B) The probability of success is the same for each observation.
C) The observations must be dependent on each other.
D) There are only two possible outcomes from each observation.
4. A lake sample was analyzed under a microscope to determine the number of clumps of alga
cells per microscope field. It was found that on average there were three clumps. What is the
expected number of clumps in four microscope fields?
A) 3
B) 6
C) 12
D) 24
5. A couple decides that they will keep having children until they have a boy. Successive births
are independent of each other and the probability of having a boy is approximately 0.50.
Does this follow a binomial setting?
A) Yes, each child has a 50% chance of being a boy.
B) Yes, each child is independent of each other.
C) No, the couple did not fix the number of observations.
6 - 7. In a matched pairs experiment, subjects pushed a button as quickly as they could after
taking a caffeine pill and also after taking a placebo pill. The mean pushes per minute were 283
for the placebo and 311 for caffeine.
6. The numbers 283 and 311 are:
A) statistics
B) parameters
C) confounded
D) random numbers
7. The parameter values in this example are:
A) 283 and 311
B) values less than 283 and greater than 311
C) two values between 283 and 311
D) unknown
8. The distribution of the averages calculated for repeated samples of size 10 of two-year-old
male children is called
A) a sampling distribution
B) a population distribution
C) a statistical distribution
D) a parametric distribution
9. The gypsy moth is a serious threat to oak trees. Traps are placed throughout the state to
detect moths. When checked, the traps contain varying numbers of moths, averaging 0.5 moths
per trap. The distribution for the individual counts is strongly skewed with a standard deviation
of 0.7. How would you describe the distribution for average number of moths per trap for 49
traps?
A) symmetrical with mean = 0.5 and std dev = 0.7
B) symmetrical with mean = 0.5 and std dev = 0.1
C) skewed with mean = 0.5 and std dev = 0.1
D) skewed with mean = 0.5 and std dev = 0.7
10. If the circumference for women’s heads is normally distributed with a mean of 22.2 and
standard deviation of 1.4 inches, what proportion of women cannot use the ready-made
helmets that exist for men due to the fact that circumferences of their heads are too small?
Ready-made helmets range from 20.24 to 25.36.
A) 0.0808
B) 0.9192
C) 0.9500
D) 0.1616
20.24 − 22.2
= −1.4
1.4
𝑃(𝑍 < −1.4) = 0.0808
11 - 13. The population of men with localized prostate tumor and a certain pretreatment has
been studied. It was found that the probability of surviving at least 5 years is equal to 0.8.
Consider a sample of 10 men with a localized prostate tumor and a pretreatment. Let X denote
the number of these patients who will survive at least 5 years.
11. The probability that at least nine patients will survive at least 5 years is:
A) 0.6778
B) 0.2020
C) 0.3757
D) 0.8242
10
10
) (.8)9 (.2)1 + ( ) (.8)10 (.2)0
9
10
= 0.2684 + 0.1073 = 0.3757
𝑃(𝑋 ≥ 9) = 𝑃(𝑋 = 9) + 𝑃(𝑋 = 10) = (
12. The probability that all will survive at least 5 years is:
A) 0.6778
B) 0.8927
C) 0.3222
D) 0.1073
10
) (.8)10 (.2)0 = .1073
10
𝑃(𝑋 = 10) = (
13. The variance of the number of patients that will survive at least 5 years in a samples of n =
10 is:
A) 1.600
B) 2.600
C) 1.265
D) 8.600
10*.8*.2=1.6
14. As part of a promotion for a new type of cracker, free trial samples are offered to shoppers
in a local supermarket. The probability that a shopper will buy a packet of crackers after
tasting the free sample is 0.200. Different shoppers can be regarded as independent trials.
If X is the number among the next 100 shoppers who buy a packet of crackers after tasting a
free sample, what is the approximate probability that at least 20 of the next 100 shoppers
who sample the crackers will buy a pack? Hint: Compute the mean and the standard
deviation, and then apply continuity-corrected normal approximation.
A) 15%
B) 25%
C) 75%
D) 55%
19.5 − 20
𝑋 − 20
𝑃(20 ≤ 𝑋) = 𝑃(19.5 ≤ 𝑋) = 𝑃 (
≤
) = 𝑃(−0.125 ≤ 𝑍)
√100(. 2)(.8) √100(. 2)(.8)
≅ 𝑃(−0.13 ≤ 𝑍) = 𝑃(𝑍 < 0.13) = 0.5517
15 - 16. A hospital delivers an average of 268 children per month. In the United States, one in
every 500 babies is born with one or more extra fingers or toes. Let X be the count of babies
born with one or more extra fingers or toes in a month at that hospital.
15. What is the variance of the number of babies born at that hospital in a month with an extra
finger or toe?
A) 0.002
B) 0.004
C) 0.536
D) 0.732
Note that Poisson random variable’s mean and variance are the same.
1
𝜇 = 268 ∗ (500) = 0.536 = 𝜎 2
16. What is the probability that at least two babies will be born with an extra finger or toe at
that hospital in a month? Hint: 𝑃(𝑋 ≥ 2)? Use the formula: 𝑃(𝑋 = 𝑘) =
figure out μ and k in this formula.
𝑒 −𝜇 𝜇 𝑘
𝑘!
You need to
A) 0.899
B) 0.314
C) 0.084
D) 0.101
𝑒 −.536 . 5360 𝑒 −.536 . 5361
[𝑃(𝑋
𝑃(𝑋 ≥ 2) = 1 −
= 0) + 𝑃(𝑋 = 1)] = 1 − [
+
]
0!
1!
= 1 − (0.585 + 0.314) = 1 − 0.899 = 0.101
17. Shelia’s measured glucose level one hour after ingesting a sugary drink varies according to
the Normal distribution with µ = 125 mg/dl and σ = 10 mg/dl. We took 16 separate
measurements from Shelia. What is the blood glucose level L such that the probability is only
0.025 that the average of 16 measurements is larger than L?
A) 129.90
B) 133.23
C) 134.80
D) 140.00
𝜎𝑋̅ =
10
= 2.5
√16
L is the 97.5th percentile of the sampling distribution of the sample mean. Since we know the
mean of the sample mean is the population mean, 125, and the standard deviation of the sample
mean is 5, L must be 125 + 1.96*2.5 =129.9, where 1.96 is the 97.5th percentile of the standard
normal distribution.
18. An insurance company knows that, in the entire population of millions of insured workers,
the mean annual cost of workers’ compensation claims is µ= $439 per insured worker, and the
standard deviation is σ = $20,000. The distribution of losses is strongly right-skewed: Most
policies have no loss, but a few have large losses, up to millions of dollars. If the company sells
100 policies, what is the probability that the mean claim loss would be no greater than $500?
A) 0.4120
B) 0.6217
C) 0.5120
D) 0.7717
𝜎𝑋̅ =
20000
(500 − 439)
= .0305
2000
√100
P(Z≤.0305) ≈ P(Z≤.03) = 0.5120
= 2000; 𝑧 =
19 - 20. 37% of the American population has blood type O+.
19. What is the probability that more than 37 out of the next 100 randomly sampled
Americans will have blood type O+?
A) 0.85
B) 0.50
C) 0.35
D) 0.25
Standard deviation of the sample proportion is equal to
√. 37 ∗ .63/100 = 0.04828043
Sample proportion, .37 = 37/100, is z = (.37-.37)/0.04828043 = 0 standard deviation up from the
mean proportion, .37, which implies P(Z>0) = 0.5 due to symmetry.
20. What is the probability that more than 370 out of the next 1000 randomly sampled
Americans will have blood type O+?
A) 0.5000
B) 0.7775
C) 0.1117
D) 0.0005
Standard deviation of the sample proportion is equal to
√. 37 ∗ .63/1000 = 0.01526761
Sample proportion, .37 = 370/1000, is (.37-.37)/0.01526761 = 0 standard deviation up from the
mean proportion, .37.
From the normal probability table, P(Z>0) = 0.5 due to symmetry.