Math 365 - Sample Exam 2
SHOW YOUR WORK on Problems 4, 6, and 7. These problems require a solution (with an
explanation) as well as an answer. A correct, but unsubstantiated answer on these problems is
worth only one point.
For Problems 1, 2, 3, and 5, fill in the blanks; no partial credit given for these problems.
Do not use books or notes. BOX your answers. You will need your graphing calculator on some
of the problems, and you are encouraged to use your calculator to verify your work. Calculator
answers must be accurate to 3 decimal places.
1. A pair of dice is rolled. Let the random variable X be the sum of the two dice. The probability
function is presented below.
X
2
3
4
5
6
7
8
9
10
11
12
prob
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Find:
(c) σ 2 =
(b) µ =
(a) P(X = 7 or X = 11) =
.
2. A continuous random variable X has the distribution shown below. Find:
(a) P(X ≥ 1) =
.
(c) P(X = 1) =
.
(b) P(-2 ≤ X < 1) =
.
3. Let X be a random variable for a binomial distribution b(8,.42). Find:
(a) µ =
(c) P(X = 6) ≈
.
(b) σ 2 =
.
.
(d) P(4 < X ≤ 7) ≈
.
4. A pediatrician reported that 30 percent of the children in the U.S. have above normal cholesterol
levels. If this is true, find the probability that in a sample of fourteen children tested for cholesterol,
more than six will have above nomal cholesterol levels.
1
5. Let Z denote the standard N(0,1)–normal distribution. Find:
(a) Find a b such that P(−b < Z < b) = .9010. b ≈
(b) P(Z ≤ .25) ≈
.
.
(c) P(Z > -1.23) ≈
.
6. The diameter of a bolt produced by the Jayhawk Machine Company is normally distributed with
a mean diameter of 0.82 cm and a standard deviation of 0.004 cm. What percent of the bolts will
meet the specification that they be between 0.816 cm and 0.826 cm?
1
7. In the United States, of the people are left–handed. In a small town (a random sample) of
6
612 persons, estimate the probability that the number of left–handed persons is strictly between
90 and 150. First find the mean (µ) and the variance (σ 2 ) where X is random variable for the
1
b(612, )–binomial distribution and use a normal distribution to approximate the probability.
6
Sample Exam No. 2 - Solutions
noindent 1. (a) P(X = 7 or X = 11) =
6
2
2
+
=
36 36
9
1
1
1
2. (a) P(X ≥ 1) = ( )(1)( ) =
2
3
6
(b) P(-2 ≤ X < 1) = 1 −
(b) µ = 7
(c) σ 2 =
1
5
=
6
6
35
.
6
(c) P(X = 1) = 0 .
3. (a) µ = 8(.42) = 3.36 . (b) σ 2 = 8(.42)(.58) = 1.9488 .
8
(c) P(X = 6) =
(.42)6 (.58)2 ≈ .0517 .
6
8
8
8
(d) P(4 < X ≤ 7) =
(.42)5 (.58)3 +
(.42)6 (.58)2 +
(.42)7 (.58) ≈ .2052 .
5
6
7
4. This is a binomial distribution b(14,.3). P(X> 6) ≈ .0933 .
5. (a) P(−b < Z < b) = .9010. So, P( Z < b) = .5 +
(b) P(Z ≤ .25) ≈ .5987
.9010
= .9505. b ≈ 1.65 .
2
(c) P(Z > -1.23) ≈ .8907 .
6. This is a normal distribution N(.82,.004).
.816 − .82
.826 − .82
≤Z≤
) ≈ .7745. Answer: ≈ 77.45%
.004
.004
1
1 5
7. µ = (612)( ) = 102
σ 2 = (612)( )( ) = 85 .
6
6 6
149.5 − 102
90.5 − 102
√
P (90 < X < 150) = P (90.5 ≤ X ≤ 149.5) ≈ P ( √
≤Z≤
) ≈ .8939 .
85
85
P (.816 < X < .826) = P (
2