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Practice Problems
1. A city’s temperature (measured at 12:00 noon) is modeled as a normal random variable with
mean and standard deviation both equal to 10 degrees Celsius. If the temperature is recorded
at 12:00 every day for one week what is the probability that the average temperature of the
measurements will be less than or equal to 11 degrees Celsius? State any assumption you
make or any approximations you are using.
I assume here that the temperature on one day is independent of any other day. Since it
is assumed that the temperature is a normal random variable on each day then the average
10
temperature for seven days is also a normal random variable with µX = 10 and σX = √
7
Thus we have that:
!
!
√ !
11 − 10
1
X − 10
7
≈ .6026
P X ≤ 11 = P
≤
= Φ 10 = Φ
10
10
√
√
√
10
7
7
7
2. Let X1 , X2 , and X3 be independent lifetimes of memory chips. Suppose that each Xi has a
normal distribution with mean 300 hours and standard deviation 10 hours.
(a) Compute the probability that at least one of the three chips lasts at least 290 hours.
To solve this problem we begin by defining the variable Y which is equal to the number
of chips of 3 which last at least 290 hours. Y is a binomial random variable with n = 3
and p = P (Xi > 290). We calculate p by:
Xi − 300
−10
p = P (Xi > 290) = P
>
= 1 − Φ(−1) = 1 − .1587 = .8413
10
10
Now, using this value p we can compute:
P (Y ≥ 1) = 1 − P (Y = 0) = 1 − p0 (1 − p)3 = 1 − (.1587)3 = .9960
(b) Compute the probability that the average lifetime of the chips is at least 290 hours.
Since the Xi ’s are identical, independent normal random variables then
√ the average of
the three IS a normal random variable and µX = 300 and σX = 10/ 3 ≈ 5.7735. We
then can compute:
!
√ X − 300
290 − 300
P (X ≥ 290) = P
≥
=
1
−
Φ
− 3 = 1 − .0418 = .9582
10
10
√
3
√
3
3. Suppose that a fair coin is tossed 900 times. Determine the probability of obtaining more
than 495 heads. State any assumptions you make or any approximations you are using.
One could do this problem in several ways. One way would be to recognize that if we define
the random variable X as the number of heads out of 900 flips. Then X is a binomial random
variable with n = 900 and p = .5. Note that np = 450 and that n(1 − p) = 450 are both
greater than 10, so we canp
approximate X with a normal random variable having mean 450
and standard deviation of np(1 − p) = 15 . Thus,
45.5
X − 450
P (X > 495) = P (X > 495.5) = P
>
= 1 − Φ(3.03) = 1 − .9988 = .0012
15
15
4. Each minute a machine produces a length of rope with mean of 4 feet and standard deviation
of 5 inches. Assuming that the amounts produced in different minutes are independent and
identically distributed, approximate the probability that the machine will produce at least
250 feet in one hour.
Let Xi for i = 1, 2, . . . 60 be the length produced during the ith minute of a given hour. The
expected value of Xi is 4 feet and the standard deviation is 5/12 feet. We then are interested
in:
P (X1 + X2 + . . . X60 ≥ 250) = P (X ≥ 250/60)
Since it is assumed that Xi is independent and identically distributed and n = 60 > 30 then
one can apply the Central Limit Theorem which says that X is approximately normal. This
gives,
!
X −4
4.16667 − 4
P (X ≥ 250/60) = P
≥
≈ 1 − Φ(3.1) = 1 − .999 = .0001
5
5
√
12 60
√
12 60
5. Suppose that the distribution of the number of defects on any given bolt of cloth is a Poisson
distribution with a mean 5, and the number of defects on each bolt is counted for a random
sample of 125 bolts. Determine the probability that the average number of defects per bolt
in the sample will be less than 5.5.
Suppose that Xi is the number of defects on the ith bolt. Each Xi is a Poisson random variable
with λ√= 5 (Since the mean is 5 which is equal to λ). The standard deviation for each Xi is
√
λ = 5. If we assume that the number of defects on each bolt is independent of the number
on any other bolt then the Xi ’s are a simple random sample with n = 125 > 30 and thus the
Central Limit Theorem will apply and the average√number of bolts is approximately normal
5
with µX = 5 and the standard deviation is σX = √125
= 15 . We then have:
P (X ≤ 5.5) = P
X −5
1
5
≤
5.5 − 5
1
5
!
≈Φ
1
2
1
5
!
5
= Φ(2.5) = .9938
=Φ
2
6. Suppose that people attending a party pour drinks from a bottle containing 63 ounces of a
certain liquid. Suppose also that the expected size of each drink is 2 ounces, that the standard
deviation of each drink is 1/2 ounce, and that all drinks are poured independently. Determine
the probability that the bottle will not be empty after 36 drinks have been poured.
Let Xi be the amount poured in each drink. The common mean is 2 ounces and the standard
deviation is .5 ounces. We seek to find:
63
P (X1 + X2 + . . . + X36 ≤ 63) = P X ≤
36
Since it is assumed that the Xi ’s are all independent and identical and the fact that n = 36 >
30 then the Central Limit Theorem applies and X is approximately normal with µX = 2 and
1
σX = √.536 = 12
. Thus we have that:
!
63
X −2
1.75 − 2
=P
≈ P (Z ≤ −3.0) = Φ(−3.0) = .0013
P X≤
≤
1
1
36
12
12
7. Assume that the helium porosity of coal samples taken from any particular seam is normally
dstributed with a true standard deviation of .75.
(a) Compute the 95% CI for the average porosity of a certain seam if the average for a
sample of 20 specimens was 4.85.
First we find η such that Φ(−η) = (1. − 95)/2 = .025. Looking at the table, η = 1.96.
Thus the CI
σ
.75
x ± η √ = 4.85 ± 1.96 √ = 4.85 ± .3287.
n
20
The interval (4.521, 5.179) is a confidence interval for the population mean µ with an
accuracy of 95%.
(b) Compute the 98% CI for the average porosity based on 16 specimens with a sample
average porosity 4.56.
First we find η such that Φ(−η) = (1. − 98)/2 = .01. Looking at the table, η = 2.33.
Thus the CI
σ
.75
x ± η √ = 4.56 ± 2.33 √ = 4.56 ± .4368.
n
16
The interval (4.123, 4.997) is a confidence interval for the population mean µ with an
accuracy of 98%.
(c) How large a sample size is necessary if the width of the 95 % CI is to be 0.4?
To solve this we set:
.75
2(1.96) √ = .4
n
.75 √
2(1.96)
= n
.4
Thus n = 54.0225. More specifically in order to have the width of the 95% confidence
interval n must be at least 55.
(d) What sample size is necessary to estimate the true average porosity to within 0.2 with
99% confidence?
To start this problem we must find η such that Φ(−η) = .005. Looking at the table
η = 2.575. Thus the maximum distance that the number can be from center of the CI
is given by:
σ
η√
n
Replacing the appropriate vales we obtain:
√
2.575(.75)/ n ≤ .2
Or,
2.575(.75)/.2 ≤
√
n
or
n ≥ 93.243
Thus, the first n that will satisfy this condition is 94.
8. Suppose that a random sample of 25 clouds are seeded with silver nitrate to see if they produce
more rain than other clouds. The rainfall from these clouds is collected and the log of these
measurements is recorded as x1 , x2 , . . . x25 . If any Xi is assumed to be normally distributed
with a common standard deviation of .3277, what is the 92% CI if the measured sample mean
is 1.379?
Since the Xi ’s are normal then the confidence interval of with 93% accuracy is given by:
.3277
σ
= 1.379 ± (1.75) √
= 1.379 ± .1147
x±η √
n
25
Thus the interval (1.2643, 1.4937) is a 93% confidence interval for µ.
9. Consider the data on the following page: Using the fact that,
n = 50,
50
X
i=1
xi = 4361.8,
50
X
(xi )2 = 414678.92
i=1
answer the following questions.
(a) What is a CI for the population mean µ with an approximate level of accuracy of 93%?
Solution: To begin we compute
p the mean as: x = 4361.8/50 = 87.236 and the sample
standard deviation as s = (414678.92 − (4361.82 /50))/(49) ≈ 26.408. The we find η
such that Φ(η) = (1 − .93)/2 = .035. Looking at the table we see that −η ≈ −1.815
Thus we have a confidence interval of
s
26.408
= 87.236 ± 6.778
x ± η √ = 87.236 ± (1.815) √
n
50
Thus the interval (80.458, 94.014) is a confidence interval for µ with accuracy of approximately 93%.
(b) What is an upper confidence bound for µ with an approximate level of accuracy 94%?
This time we look up η such that Φ(−η) = 1 − .94 = .06 which gives η ≈ 1.55. Thus an
upper confidence bound is:
s
26.408
µ<x+η √
= 87.236 + (1.55) √
= 93.0247
n
50
is an upper confidence bound with approximate accuracy of 94%.
(c) Make an estimate of what number of measurements should be taken to assure that the
width of a CI with 90% accuracy is no more than 1.2.
For this problem σ is not known. So since we are trying to guess n so we must
come up with a guess for s. For this we use (max − min)/4 which is equal to (113.5 −
5.3)/4 = 27.05. We also look up η such that Φ(−η) = .05 which gives −η ≈ −1.645.
The width of the confidence interval with approximately 90% accuracy is given by 2 ∗
√
(1.645)(27.05)/ n. Setting this less than 1.2 and solving for n gives:
√
2 ∗ (1.645)(27.05)/ n < 1.2
√
2 ∗ (1.645)(27.05)/1.2 < n
(2 ∗ (1.645)(27.05)/1.2)2 < n
Data Series:
5.3, 6, 8.2, 13.8, 40.8, 52.6, 74.1, 79, 85.3, 88, 90.2, 91.2, 91.5, 92.4, 92.5, 93.1, 93.6, 94.1, 94.1,
94.3, 94.3, 94.4, 94.8, 94.8, 94.8, 94.9, 95.2, 95.5, 95.7, 95.8, 95.8, 95.9, 96.3, 96.6, 96.7, 97.8,
98.1, 98.5, 99, 99.1, 99.8, 100.7, 101.4, 102.7, 103.7, 106, 107.6, 110.2, 112.1, 113.5,