Probability & Statistics/Spring 2023/Review #3
Exam is on Monday, May. 8, 11:45am-2:45pm, 2023
You may use a calculator. You may also bring in a single sheet of 8 12 × 1100
paper with anything you want written on it.
You will be asked questions similar to those on quizzes, and homeworks, and to
examples done in class and in notes.
You will be supplied with normal distribution tables, χ-square tables, and ttables.
Topics
The exam will be based on the material covered in class with a similar emphasis.
The relevant sections of the textbook are:
Chapter 1: 1.1–1.7
Chapter 2: 2.1–2.6
Chapter 3: 3.1
Chapter 5: 5.1–5.3
Chapter 7: 7.1–7.3
Chapter 8: 8.1–8.2
Chapter 9: 9.1–9.3
Chapter 10: 10.3–10.4
Chapter 12: 12.1–12.3
Sample problems
The following sample problems are not intended to provide an exhaustive review
of the course, but to remind you of the major themes.
1. Two dice are thrown and the scores on each recorded. Let A be the event
that the first die is even, and B be the event that the sum of the dice is
odd. Are A and B
(a) mutually exclusive?
(b) independent?
2. How many 5-card poker hands contain a pair of eights and a pair of aces?
3. The events B1 and B2 constitute a partition of the sample space S, so
B1 ∩ B2 = ∅, B1 ∪ B2 = S. An event A satisfies P (A|B1 ) = 41 , P (A|B2 ) =
1
1
2 . Given that P (B1 ) = 3 , find P (A) and P (B1 |A).
4. X1 and X2 are independent indentically distributed random variables with
µ = 2, σ 2 = 3. Find the mean and variance of
(a) X1 + X2
(b) X1 − X2
(c) X1 + 3X2
5. (a) Let X be the number of heads when a fair coin is tossed 10 times.
Find P (4 ≤ X ≤ 6).
(b) Let X be the number of heads when a fair coin is tossed 100 times.
Approximate P (40 ≤ X ≤ 60).
6. Let Z denote the standard normal distribution, and X the normal distribution with µ = 4, σ = 2. Find
(a) P (−1 ≤ Z ≤ 3).
(b) P (3 < X < 5).
(c) P (X > 2.5).
7. Suppose X is a random variable with σ = 2. If a sample of size 100 has a
mean of 50, give a 95% confidence interval for µ = E(X).
8. A coin is tossed 10 times and comes up heads 9 times. Using a 5% level
of significance, would you conclude the coin is biased towards heads?
9. A manufacturer claims that its light bulbs have a life of at least 1000
hours. A random sample of 16 light bulbs has a mean life of 1030 hours
with a standard deviation of 50 hours. Assuming that the light bulb lives
are normally distributed, and using α = 0.05 is the manufacturer’s claim
reasonable?
P
10. Consider
n = 40,
x=
P the following
P 2 data forPsimple linear
Pregression:
120,
y = 200,
x = 1000,
xy = 500,
y 2 = 2000.
(a) Find the regression coefficients β̂0 and β̂1 .
(b) Using α = 0.05, test the hypotheses H0 : β1 = 0 versus H1 : β1 6= 0.
P
11. P
Consider the following
simple
linear regressionPdata: n = 11,
x = 99,
P
P
x2 = 1001,
y = 82.51,
y 2 = 660.1727,
xy = 797.6.
(a) Find x̄ and ȳ.
(b) Find Sxx , Sxy and Syy .
(c) Find the regression coefficients β̂0 and β̂1 .
(d) Estimate the value of y if x = 9.
(e) Find an estimate σ̂ 2 for the variance.
(f) Using a 5% significance level, test the hypotheses H0 : β1 = 0 versus
H1 : β1 6= 0.
Solutions to sample problems
1. (a) No: for example (2, 1) ∈ A ∩ B.
(b) Yes: P (A) = 12 , P (B) = 21 , P (A ∩ B) =
2. 42 42 44
1 = 6 × 6 × 44 = 1584.
3. P (A) = P (A|B1 )P (B1 )+P (A|B2 )P (B2 ) =
1
5.
1
4
5
12 ,
= P (A)P (B).
P (B1 |A) =
P (A|B1 )P (B1 )
P (A)
=
4. (a) µ = 4, σ 2 = 6.
(b) µ = 0, σ 2 = 6.
(c) µ = 8, σ 2 = 30.
5. (a)
P (4 ≤ X ≤ 6) = P (X = 4) + P (X = 5) + P (X = 6)
10 10 10
10
1
10
1
10
1
+
+
=
2
5
2
6
2
4
672
= 10 = 0.65625.
2
(b) We use the normal approximation to the binomial. X ∼ B(100, 12 ),
so µ = 50, σ 2 = 25. Let Y ∼ N (µ, σ 2 ), Z ∼ N (0, 1). Then
P (40 ≤ X ≤ 60) ≈ P (39.5 ≤ Y ≤ 60.5)
10.5
10.5
=P −
≤Z≤
5
5
= P (−2.1 ≤ Z ≤ 2.1)
= 0.9642.
6. (a) 0.8400
(b) 0.3830
(c) 0.7734
7. z0.025 = 1.96, interval is [50 − 1.96 ×
2
2
10 , 50 + 1.96 × 10 ]
= [49.608, 50.392].
8. Let p be the probability of getting a head on each toss, and let X be the
number of heads. Then X ∼ N (10, p, We want to test the hypotheses
H0 : p ≤ 12 versus H1 : p > 21 . Suppose the null hypothesis is true. We
calculate the probability of 9 or more heads as P (X ≥ 9) = 211
10 ≈ 0.0107.
This is less than 5%, so we would reject H0 in favor of H1 , and conclude
that the coin is biased towards heads.
9. We are testing H0 : µ ≤ µ0 versus H1 : µ > µ0 , where µ0 = 1000. Then
√ 0 = 2.4, and t0.05,15 = 1.753, so we reject H0 in favor of H1 . The
t = x̄−µ
s/ n
manufacturer’s claim is reasonable.
10. We calculate x̄ = 3, ȳ = 5, Sxx = 640, Sxy = −100, Syy = 1000.
(a) β̂1 = Sxy /Sxx = −0.15625, β̂0 = ȳ − x̄β̂1 = 5.46875.
(b) Using α = 0.05, test the hypotheses H0 : β1 = 0 versus H1 : β1 6= 0.
We estimate the variance as σ̂ 2 =
Syy −β̂12 Sxx
n−2
≈ 25.905. Then t =
β̂1
√
σ̂/ Sxx
= −0.7766, t38,0.025 = 2.024, so we fail to reject H0 , and
cannot conclude β1 6= 0.
P
P
11. (a) x̄ = n1
x = 9, ȳ = n1
y = 7.5.
P 2
P
P 2
2
(b) Sxx =
x − nx̄ = 110, Sxy =
xy − nx̄ȳ = 55.1, Syy =
y −
nȳ 2 = 41.42.
(c) β̂1 = Sxy /Sxx = 0.5, β̂0 = ȳ − x̄β̂1 = 3.
(d) y(9) ≈ β̂0 + 9β̂1 = 7.5.
(e) σ̂ 2 =
Syy − β̂12 Sxx
≈ 1.54697.
n−2
β̂
√1
= 4.216. Since t9,0.025 = 2.262, we reject H0 in favor of
σ̂/ Sxx
H1 and conclude β1 6= 0.
(f) t =