STT 041.1 Problem Set #2 - Binomial Distribution
1. A multiple-choice quiz has 15 questions, each with 5 possible answers of which
only one is the correct answer. What is the probability that sheer guesswork
yields
a) from 5 to 10 correct answers?
There are total 15 multiple choice questions.
P(C) = probability of answering any question correctly = (1 /5 ) = 0.20
P(W) = probability of answering any question wrongly = (4 / 5) = 0.80
Probability of answering 5 to 10 questions correctly = Adding
Probability of answering exactly 5 questions correctly to Probability of
answering exactly 10 questions correctly
P(5)= {(15C5) * (0.20^5) * (0.80^10)} = 0.1032
P(6)={(15C6) * (0.20^6) * (0.80^9)} = 0.0430
P(7) = {(15C7) * (0.20^7) * (0.80^8)} = 0.0138
P(8)= {(15C8) * (0.20^8) * (0.80^7)} = 0.0035
P(9) = {(15C9) * (0.20^9) * (0.80^6)} = 0.0006
P(10) = {(15C10) * (0.20^10) * (0.80^5)} = 0.0001
≈ (0.1032 + 0.0429 +0.0138+ 0.0034 + 0.0006 + 0.0001)
≈ 0.1642
b) Exactly 2 correct answers?
n = 15, p = (1/5) = .20
P(x=2) = (15C2)* (.20^2)*(.80*13)
= 0.2309
c) From 3 to 4 correct answers? 0.75
N = 15
P(C) = probability of answering any question correctly = (1 /5 ) = 0.20
P(W) = probability of answering any question wrongly = (4 / 5) = 0.80
P(3)= {(15C3) * (0.20^3) * (0.80^12)} = 0.2501
P(4)={(15C4) * (0.20^4) * (0.80^11)} = 0.1876
≈ (0.2501 + 0.1876)
≈ 0.4377
d) At least 5 correct answers?
n
P(x=r) = pr (1 − p)n−r
r
n = 15, p=(1/5) = .20 q=(4/5)=.80
P (x ≥ 5) = 1- (p < 5)
e) At most 3 correct answers?
n r
p (1 − p)n−r
r
n = 15, p=(1/5) = .20 q=(4/5)=.80
P(x=r) =
P (≤ 3)
P(0)= {(15C0) * (0.20^0) * (0.80^15)} = 0.0351
P(1)={(15C1) * (0.20^1) * (0.80^14)} = 0.1319
P(2) = {(15C2) * (0.20^2) * (0.80^13)} = 0.2309
P(3)= {(15C3) * (0.20^3) * (0.80^12)} = 0.2501
≈ (0.0351+0.1319+0.2309+0.2501)
≈ 0.648
2. Visiting the Dentist. Sixty-three percent of adults say they are visiting the
dentist less because of the economy. You randomly select six adults and ask
them if they are visiting the dentist less because of the economy.
(Source: American
Optometric Association)
a. construct a binomial distribution,
p = 0.63, q = 0.37, n = 6
P(0) = {(6C0) * (0.63^0) * (0.37^6)} = 0.0003
P(1) = {(6C1) * (0.63^1) * (0.37^5)} = 0.0262
P(2) = {(6C2) * (0.63^2) * (0.37^4)} = 0.1116
P(3) = {(6C3) * (0.63^3) * (0.37^3)} = 0.2533
P(4) = {(6C4) * (0.63^4) * (0.37^2)} = 0.3234
P(5) = {(6C5) * (0.63^5) * (0.37^1)} = 0.2203
P(6) = {6C6) * (0.63^6) * (0.37^0)} = 0.0625
≈ (0.0003 + 0.0262 +0.1116+ 0.2533 + 0.3234 + 0.2203 +0.0625)
≈ 0.9976
b. graph the binomial distribution using a histogram and describe its shape,
The graph shows a skewed shape.
c. find the mean, variance, and standard deviation of the binomial
distribution, and
Mean:
m = np = 6*.63
≈ 3.78
Variance:
o2 = npq = 6 ∗ .63 ∗ .37
≈ 1.3986
Standard Deviation:
o =√npq = √6 ∗ .63 ∗ .37
≈ 1.1826
d. interpret the results in the context of the real-life situation. What values
of the random variable x would you consider unusual? Explain your
reasoning.
From the histogram, you can see that it would be unusual if none
or zero, only one, or all six of the adults visit the dentist less for
adults can’t afford such dental fees because of the low economy.
Putting in a situation where less developed countries can’t really
afford their food how much more to the health-related matters just
like these dental matters.