JustAnswer
Questions (answers are all below):
36. Dr. Stallter has been teaching basic statistics for many years. She knows that 80 percent of the students will complete the assigned problems. She has also determined that among those who do their assignments, 90 percent will pass the course. Among those students who do not do their homework, 60 percent will pass. Mike Fishbaugh took statistics last semester from Dr. Stallter and received a passing grade. What is the probability that he completed the assignments?
38. One-fourth of the residents of the Burning Ridge Estates leave their garage doors open when
they are away from home. The local chief of police estimates that 5 percent of the garages
with open doors will have something stolen, but only 1 percent of those closed will have
something stolen. If a garage is robbed, what is the probability the doors were left open?
45. A Tamiami shearing machine is producing 10 percent defective pieces, which is abnormally
high. The quality control engineer has been checking the output by almost continuous
sampling since the abnormal condition began. What is the probability that in a sample of
10 pieces:
a. Exactly 5 will be defective?
b. 5 or more will be defective?
62. Suppose 1.5 percent of the antennas on new Nokia cell phones are defective. For a random
sample of 200 antennas, find the probability that:
a. None of the antennas is defective.
b. Three or more of the antennas are defective.
42. The accounting department at Weston Materials, Inc., a national manufacturer of unattached
garages, reports that it takes two construction workers a mean of 32 hours and
a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times
follow the normal distribution.
a. Determine the z values for 29 and 34 hours. What percent of the garages take between
32 hours and 34 hours to erect?
b. What percent of the garages take between 29 hours and 34 hours to erect?
c. What percent of the garages take 28.7 hours or less to erect?
d. Of the garages, 5 percent take how many hours or more to erect?
45. Shaver Manufacturing, Inc., offers dental insurance to its employees. A recent study by the
human resource director shows the annual cost per employee per year followed the normal
probability distribution, with a mean of $1,280 and a standard deviation of $420 per year.
a. What fraction of the employees cost more than $1,500 per year for dental expenses?
b. What fraction of the employees cost between $1,500 and $2,000 per year?
c. Estimate the percent that did not have any dental expense.
d. What was the cost for the 10 percent of employees who incurred the highest dental
expense?
ANSWERS:
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Stallter:
P = .80*.90 / (.20*.60 + .80*.90)
(calculator)
P = 0.857
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Burning Ridge:
We need:
prob(open | robbed)
Using Bayes Theorem:
= prob(open)* prob(robbed | open) / (prob(open)*prob(robbed | open) + prob(not open)*prob(robbed | closed))
Plug in our numbers:
= 0.25*0.05 / (0.25*0.05 + 0.75*0.01)
= 0.625 (which is 5/8)
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Tamiami
This uses the binomial formula.
prob(x) = N choose X * p^x * (1-p)^(N-x)
N = 10
p = 0.1
1-p = 0.9
So:
Exactly 5:
10 choose 5 * 0.1^5 * 0.9^5
= 252 * 0.1^5 * 0.9^5
= 0.0014880348
5 or more:
10 choose 5 * 0.1^5 * 0.9^5 + 10 choose 6 * 0.1^6 * 0.9^4 + 10 choose 7 * 0.1^7 * 0.9^3 + 10 choose 8 * 0.1^8 * 0.9^2 + 10 choose 9 * 0.1^9 * 0.9^1 + 10 choose 10 * 0.1^10 * 0.9^0
= 252 * 0.1^5 * 0.9^5 + 210 * 0.1^6 * 0.9^4 + 120 * 0.1^7 * 0.9^3 + 45 * 0.1^8 * 0.9^2 + 10 * 0.1^9 * 0.9^1 + 1 * 0.1^10 * 0.9^0
= 0.0016349374
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Nokia
Using the binomial formula:
-None of the antennas is defective.
= (1 - 0.015)^200
= about 0.04866829
-Three or more of the antennas are defective.
= 1 - zero defective - one defective - two defective
= 1 - (1 - 0.015)^200 - 200*(0.015)^1*(1-0.015)^(199) - 19900*(0.015)^2*(1-0.015)^(198)
= about 0.5785
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The z score tells you how many standard deviations above/below the mean your sample is.
The formula is:
z = (x-mu)/sigma
In your problem, mu = 32, sigma = 2.
z(29) = (29-32)/2 = -1.5
z(34) = (34-32)/2 = 1
What percent of the garages take between32 hours and 34 hours to erect?
using a z table ():
p(0 < z < 1) = 0.3413 = 34.13%
b. What percent of the garages take between 29 hours and 34 hours to erect?
using the z table again:
p(-1.5 < z < 1) = 0.7745 = 77.45%
c. What percent of the garages take 28.7 hours or less to erect?
Get the z value:
z = (28.7-32)/2 = -1.65
using the table:
p(z < -1.65) = 0.0495 = 4.95%
d. Of the garages, 5 percent take how many hours or more to erect?
From the table, the z value we need is:
1.6449
Solve for x:
1.6449 = (x-32)/2
Multiply by 2:
2*1.6449 = (x-32)
Add 32:
x = 2*1.6449 + 32
x = 35.2898 hours
---------------------------------
We'll need to get z scores for each of these.
z = (x - mu)/sigma, where x is the sample point, mu is the mean, and sigma is the standard deviation.
a. What fraction of the employees cost more than $1,500 per year for dental expenses?
z = (1500-1280)/420
z = 0.5238
From a z table ():
prob(z > 0.5238) = 0.3002 (about 3/10)
b. What fraction of the employees cost between $1,500 and $2,000 per year?
z(2000) = (2000-1280)/420
z = 1.71429
From the table:
prob(0.5238 < z < 1.71429) = 0.2570 (just over 1/4)
c. Estimate the percent that did not have any dental expense.
z(0) = (0-1280)/420
z = -3.0476
From the table:
prob(z < -3.0476) = 0.0012 (0.12%)
d. What was the cost for the 10 percent of employees who incurred the highest dental
expense?
From the table, we are looking for z = 1.2816. Solve for x:
z = 1.2816 = (x-1280)/420
Multiply by 420:
x-1280 = 420*1.2816
Add 1280:
x = 420*1.2816 + 1280
x = 1818.272
Greater than $1818.27
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