Download Solution to Additional Problems 1. The length of useful life

Solution to Additional Problems
1. The length of useful life a fluorescent tube used for indoor gardening is normally distributed. The useful life
has a mean of 600 hrs and a standard deviation of 40 hrs. Determine the probability that
(a) a tube chosen at random will last between 620 and 680 hrs.
P (620 < X < 680) = P ( 620−600
<
40
X−600
40
<
680−600
)
40
= P (.5 < Z < 2) = .28578
(b) such a tube will last more than 740 hrs.
P (X > 740) = P ( X−600
>
40
740−600
)
40
= P (Z > 3.5) = .00023
(c) such a tube will last less than 475 hrs.
P (X < 475) = P ( X−600
<
40
475−600
)
40
= P (Z < −3.125) = .00089
(d) a tube chosen at random will last less than 480 hrs or more than 720 hrs.
P (X < 480) + P (X > 720) = P ( X−600
<
40
480−600
)
40
+ P ( X−600
>
40
720−600
)
40
= P (Z < −3) + P (Z > 3) =
.00135 + .00135 = .0027
2. The waiting time at a certain bank is approximately normally distributed with a mean of 3.7 min and a standard
deviation of 1.4 min.
(a) Find the probability that a randomly selected customer has to wait less than 2.0 min.
Z=
2−3.7
1.4
= −1.2143. Hence, the probability is .1123 (area under the normal curve to the left of -1.2143)
(b) What is the waiting time T such that the probability that a customer waits longer than T is only 0.05?
T = Z × σ + µ = 1.645(1.4) + 3.7 = 6 mins
(c) Find the probability that a randomly selected customer has to wait more than 6.0 min.
Based on part b: probability is approximately equal to .05
(d) If a random sample of 2 customers is selected, find the probability that the average waiting time is more
than 6.0 min.
σX̄ =
1.4
√
2
= .9899, Z =
6−3.7
.9899
= 2.323. Hence, the probability is .0101 (area under the normal curve to
the right of 2.323)
(e) If a random sample of 4 customers is selected, find the probability that the average waiting time is more
than 6.0 min.
σX̄ =
1.4
√
4
= .70, Z =
6−3.7
.70
= 3.29. Hence, the probability is .0005 (area under the normal curve to the
right of 3.29)
(f) If a random sample of 10 customers is selected, find the probability that the average waiting time is more
than 6.0 min.
σX̄ =
1.4
√
10
= .4427, Z =
6−3.7
.4427
= 5.195. Hence, the probability is approximately equal to 0(area under the
normal curve to the right of 5.195)
(g) What is the waiting time T such that the probability that X̄ - average waiting time of 10 customers - is
greater than T is only 0.05?
√
√
T = Z × σ/ n + µ = 1.645(1.4/ 10) + 3.7 = 4.43 mins
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(h) A random sample of 10 customers yielded an average waiting time of 5 min., what is the probability for
observing this outcome or more extreme than the 5 min? Is this a unlikely event?
σX̄ =
1.4
√
10
= .4427, Z =
5−3.7
.4427
= 2.936. Hence, the probability is .00166 (area under the normal curve to
the right of 2.936). Yes, this is a very unlikely event.
3. According to the 1993 World Factbook, the 1993 total fertility rate (mean number of children born per woman)
for Madagascar is 6.75. Suppose the standard deviation of the total fertility rate is 2.5. The mean number of
children for a sample of 200 randomly selected women is one value of many that form the sampling distribution
of sample means.
(a) What is the shape of this sampling distribution?
By the central limit theorem (CLT), the sampling distribution of the sample means is normal (bell-shaped).
(b) What is the mean value for this sampling distribution?
The mean (µ) for this sampling distribution is equal to the population mean 6.75.
(c) What is the standard deviation of this sampling distribution?
The standard deviation of this sampling distribution is σX̄ =
√2.5
200
= .1768
4. According to an article in the January 1991 issue of Health magazine, root-canal therapy costs from $200 to
$700. Suppose the mean cost for root-canal therapy is $450 and the standard deviation is $125.
(a) If a sample of 50 dentists was selected across the country, find the probability that the mean cost per
root-canal for the sample between $425 and $475.
σX̄ =
125
√
50
= 17.68, P (425 < X < 475) = P ( 425−450
<
17.68
X−450
17.68
<
475−450
17.68 )
= P (−1.414 < Z < 1.414) =
.8426
(b) If a sample of 100 dentists was selected across the country, find the probability that the mean cost per
root-canal for the sample between $425 and $475.
σX̄ =
√125
100
= 12.5, P (425 < X < 475) = P ( 425−450
<
12.5
X−450
12.5
<
475−450
12.5 )
= P (−2 < Z < 2) = .9545
(c) If the probability that the mean cost per root-canal for the sample between $440 and $460 is .95, what
sample size is needed to satisfy these conditions? Since the population mean is right in the middle of 440
√ + 450 or 440 = −1.96 × 125
√ + 450. This is equivalent to the
and 460 we can consider: 460 = 1.96 × 125
n
n
√
√
∗
margin of error m = z σ/ n, 10 = 1.96(125)/ n. Solving for n = 601.
5. According to the USA Snapshot “Knowing drug addicts,” 45% of Americans know somebody who became
addicted to a drug other than alcohol. Assuming this to be true, what is the probability that
(a) exactly 3 of a random sample of 5 know someone who became addicted?
Binomial distribution with n = 5, x = 3, p = .45, Table gives P (X = 3) = .2757
(b) exactly 7 of a random sample of 15 know someone who became addicted?
Binomial distribution with n = 15, x = 7, p = .45, Table gives P (X = 7) = .2013
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(c) at least 7 of a random sample of 15 know someone who became addicted?
Binomial distribution with n = 15, x ≥ 7, p = .45, Table gives P (X ≥ 7) = .2013 + .1657 + .1049 + .0514 +
.0192 + .0052 + .0010 + .0001 + .0000 = .5478
(d) no more than 7 of a random sample of 15 know someone who became addicted?
Binomial distribution with n = 15, x ≤ 7, p = .45, Table gives P (X ≤ 7) = .6535
(e) at least 1000 of a random sample of 2000 know someone who became addicted?
Binomial distribution with n = 2000, x ≥ 1000, p = .45. Use the normal approximation: µ = np =
p
p
2000(.45) = 900 and σ = np(1 − p) = 2000(.45)(.55) = 22.25, Z = 1000−900
= 4.494. Hence, the
22.25
probability is .0000 (area under the normal curve to the right of 4.494.)
(f) the proportion of a random sample of 2000 know someone addicted to a drug exceeds 55%?
This is equivalent to finding the probability that at least 2000(.55) = 1100 of a random sample of 2000
know someone addicted to a drug. Binomial distribution with n = 2000, x ≥ 1100, p = .45. Use the
p
p
normal approximation: µ = np = 2000(.45) = 900 and σ = np(1 − p) = 2000(.45)(.55) = 22.25,
Z=
1100−900
22.25
= 8.99. Hence, the probability is .0000 (area under the normal curve to the right of 8.99.)
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