Download Math 227 Practice Test 3

Math 227 Practice Test 3 ( Ch 6 and 7)
Name___________________________________
Assume that the weight loss for the first month of a diet program varies between 6 pounds and 12 pounds, and is spread
evenly over the range of possibilities, so that there is a uniform distribution. Find the probability of the given range of
pounds lost.
1) Between 8.5 pounds and 10 pounds
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A)
B)
C)
D)
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Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard
deviation 1.
2)
A) 0.2224
B) 0.2190
C) 0.2776
D) 0.7224
Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.
3) Shaded area is 0.9599.
A) 1.75
B) -1.38
If z is a standard normal variable, find the probability.
4) P(z > 0.59)
A) 0.2190
B) 0.7224
C) 1.82
D) 1.03
C) 0.2776
D) 0.2224
The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0°C at
the freezing point of water. Tests on a large sample of these thermometers reveal that at the freezing point of water, some
give readings below 0°C (denoted by negative numbers) and some give readings above 0°C (denoted by positive
numbers). Assume that the mean reading is 0°C and the standard deviation of the readings is 1.00°C. Also assume that the
frequency distribution of errors closely resembles the normal distribution. A thermometer is randomly selected and
tested. Find the temperature reading corresponding to the given information.
5) If 7% of the thermometers are rejected because they have readings that are too high, but all other thermometers
are acceptable, find the temperature that separates the rejected thermometers from the others.
A) 1.26°
B) 1.45°
C) 1.39°
D) 1.48°
Find the indicated value.
6) z 0.005
A) 2.535
B) 2.835
C) 2.015
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D) 2.575
Provide an appropriate response.
7) Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of
15 (as on the Wechsler test). Find P30, which is the IQ score separating the bottom 30% from the top 70%.
A) 92.8
B) 91.9
C) 91.4
D) 92.2
Solve the problem. Round to the nearest tenth unless indicated otherwise.
8) Suppose that replacement times for washing machines are normally distributed with a mean of 8.4 years and a
standard deviation of 2 years. Find the replacement time that separates the top 18% from the bottom 82%.
A) 6.6 years
B) 10.2 years
C) 9.7 years
D) 8.8 years
Assume that X has a normal distribution, and find the indicated probability.
9) The mean is µ = 15.2 and the standard deviation is = 0.9.
Find the probability that X is between 14.3 and 16.1.
A) 0.8413
B) 0.1587
C) 0.6826
D) 0.3413
Find the indicated probability.
10) A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a
standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 200
and 275.
A) 0.5
B) 0.9332
C) 0.4332
D) 0.0668
Solve the problem.
11) The scores on a certain test are normally distributed with a mean score of 60 and a standard deviation of 5.
What is the probability that a sample of 90 students will have a mean score of at least 60.527?
A) 0.3413
B) 0.1587
C) 0.3174
D) 0.8413
12) A final exam in Math 160 has a mean of 73 with standard deviation 7.8. If 24 students are randomly selected,
find the probability that the mean of their test scores is greater than 78.
A) 0.0036
B) 0.0008
C) 0.0103
D) 0.8962
Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution.
13) A certain question on a test is answered correctly by 22% of the respondents. Estimate the probability that
among the next 150 responses there will be at most 40 correct answers.
A) 0.0694
B) 0.9306
C) 0.8997
D) 0.1003
14) The probability that a radish seed will germinate is 0.7. Estimate the probability that of 140 randomly selected
seeds, exactly 100 will germinate.
A) 0.9331
B) 0.0669
C) 0.0679
D) 0.0769
Use the normal distribution to approximate the desired probability.
15) Find the probability that in 200 tosses of a fair die, we will obtain at exactly 30 fives.
A) 0.0429
B) 0.0871
C) 0.1871
D) 0.0619
Find the indicated critical z value.
16) Find z /2 for = 0.07.
A) 2.70
B) 1.81
C) 1.96
17) Find the critical value z /2 that corresponds to a 98% confidence level.
A) 2.05
B) 1.75
C) 2.33
2
D) 1.48
D) 2.575
Express the confidence interval using the indicated format.
^
^
18) Express the confidence interval 0.491 ± 0.057 in the form of p - E < p < p + E.
A) 0.434 < p < 0.491
B) 0.434 < p < 0.548
C) 0.491 < p < 0.548
D) 0.4625 < p < 0.5195
Solve the problem.
19) The following confidence interval is obtained for a population proportion, p: (0.399, 0.437). Use these confidence
interval limits to find the margin of error, E.
A) 0.019
B) 0.017
C) 0.020
D) 0.038
Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the
given statistics and confidence level. Round the margin of error to four decimal places.
20) In a clinical test with 1600 subjects, 800 showed improvement from the treatment. Find the margin of error for
the 99% confidence interval used to estimate the population proportion.
A) 0.0245
B) 0.0282
C) 0.0184
D) 0.0322
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
21) n = 195, x = 162; 95% confidence
A) 0.789 < p < 0.873
B) 0.788 < p < 0.873
C) 0.778 < p < 0.883
D) 0.777 < p < 0.884
Use the given data to find the minimum sample size required to estimate the population proportion.
^
^
22) Margin of error: 0.008; confidence level: 98%; p and q unknown
A) 10,384
B) 21,207
C) 22,184
D) 20,308
^
23) Margin of error: 0.07; confidence level: 95%; from a prior study, p is estimated by the decimal equivalent of 92%.
A) 58
B) 4
C) 174
D) 51
Solve the problem. Round the point estimate to the nearest thousandth.
24) 32 randomly picked people were asked if they rented or owned their own home, 8 said they rented. Obtain a
point estimate of the proportion of home owners.
A) 0.750
B) 0.200
C) 0.781
D) 0.250
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
25) Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the
true proportion of all adults in the town who have health insurance.
A) 0.566 < p < 0.760
B) 0.582 < p < 0.744
C) 0.536 < p < 0.790
D) 0.548 < p < 0.778
Solve the problem.
26) In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard
deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh
more than 180 pounds? Assume that we want 96% confidence that the error is no more than 4 percentage points.
A) 658
B) 501
C) 317
D) 232
Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume
that the population has a normal distribution.
27) n = 12, x = 23.6, s = 6.6, 99% confidence
A) 17.70 < µ < 29.50
B) 18.42 < µ < 28.78
C) 17.56 < µ < 29.64
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D) 17.68 < µ < 29.52
Use the given information to find the minimum sample size required to estimate an unknown population mean µ.
28) How many business students must be randomly selected to estimate the mean monthly earnings of business
students at one college? We want 95% confidence that the sample mean is within $135 of the population mean,
and the population standard deviation is known to be $538.
A) 54
B) 86
C) 62
D) 43
Use the confidence level and sample data to find a confidence interval for estimating the population µ. Round your
answer to the same number of decimal places as the sample mean.
29) A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a
placement test. What is the 90% confidence interval for the mean score, µ, of all students taking the test?
A) 28.4 < µ < 30.6
B) 27.9 < µ < 31.1
C) 28.2 < µ < 30.8
D) 27.8 < µ < 31.2
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