1. Find the mean for the given sample data. Unless indicated otherwise, round your answer to one more decimal place than is present in the original data values.
The students in Hugh Logan's math class took the Scholastic Aptitude Test. Their math scores are shown below. Find the mean score.
536, 608, 344, 340, 596, 357, 343, 566, 470, 482
(Points : 4)
464.2
476.0
473.7
455.1
Solution: A. 464.2
2. Find the midrange for the given sample data.
Listed below are the amounts of time (in months) that the employees of an electronics company have been working at the company. Find the midrange.
12, 21, 26, 37, 46, 53, 61, 66, 75, 76, 85, 91, 132, 155
W4T6 (Points : 4)
63.5 months
71.5 months
66.9 months
83.5 months
The midrange is the average of the minimum and maximum values.
The minimum value is 12, and the maximum value is 155.
Solution: D. 83.5 months
3. Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in the original data.
20.0, 21.5, 27.4, 47.3, 13.1, 11.1
W4T7 (Points : 4)
4145.12
37.34
13.11
3285.46
Solution: C. 13.11
4. Find the indicated probability.
A bag contains 5 red marbles, 3 blue marbles, and 1 green marble. Find P(not blue). (Points : 4)
1/3
3/2
6
2/3
The total number of marbles is 5 + 3 + 1 = 9.
The number of marbles that are not blue is 5 + 1 = 6
Then P(not blue) = (number of marbles that are not blue) / (total number of marbles)
P(not blue) = 6 / 9
P(not blue) = 2/3
Alternatively, you could calculate P(blue), and then subtract that from 1 to get P(not blue):
P(blue) = (number of blue marbles) / (total number of marbles)
P(blue) = 3/9 = 1/3
P(not blue) = 1 – P(blue) = 1 – (1/3) = 2/3
5. Find the indicated probability. Round to the nearest thousandth.
In a blood testing procedure, blood samples from 6 people are combined into one mixture. The mixture will only test negative if all the individual samples are negative. If the probability that an individual sample tests positive is 0.11, what is the probability that the mixture will test positive? (Points : 4)
0.00000177
1.000
0.503
0.497
The probability of a positive test is given as P(positive) = 0.11.
The probability of a negative test, P(negative), is 1 – P(positive) = 1 – 0.11 = 0.89
The probability that all 6 samples test negative is:
P(all 6 test negative) = (0.89)6 = 0.49698
The probability that the mixture tests negative is then also 0.497
The probability that the mixture test positive is then 1 – 0.497 = 0.503
Solution: C. 0.503
6. Evaluate the expression.
11C4 (Points : 4)
330
1980
5040
3
Solution: A. 330
7. Provide an appropriate response.
Suppose you pay $3.00 to roll a fair die with the understanding that you will get back $5.00 for rolling a 1 or a 6, nothing otherwise.
What is your expected value? (Points : 4)
-$3.00
$3.00
-$1.33
$5.00
The probability of rolling a 1 or a 6, “winning”, is 2/6 = 1/3.
The probability of rolling any other number, “losing”, is 1 – 1/3 = 2/3.
Expected value = P(winning)($5) + P(losing)($0) - $3.00
Expected value = (1/3)($5) + (2/3)($0) - $3.00
Expected value = -$1.33
Solution: C. -$1.33
8. Assume that a researcher randomly selects 14 newborn babies and counts the number of girls selected, x. The probabilities corresponding to the 14 possible values of x are summarized in the given table. Answer the question using the table.
W4T15
Find the probability of selecting 2 or more girls. (Points : 4)
0.999
0.006
0.001
0.994
Without the table showing the probability distribution, this problem cannot be answered.
Here is how to go about solving it though:
Use the values in the table, and subtract the probabilities for selecting 0 girls and for selecting 1
girl from 1.00.
The result will be the probability of 2 or more girls.
9. Determine whether the given procedure results in a binomial distribution. If not, state the reason why.
Choosing 10 marbles from a box of 40 marbles (20 purple, 12 red, and 8 green) one at a time with replacement, keeping track of the number of red marbles chosen. (Points : 4)
Not binomial: there are more than two outcomes for each trial.
Procedure results in a binomial distribution.
Not binomial: the trials are not independent.
Not binomial: there are too many trials.
This scenario meets all of the criteria of a binomial experiment:
1. Each trial can have only two outcomes:
red or not red
2. There are a fixed number of trials:
10
3. The outcomes of each trial are independent:
The result of one draw does not affect the result of any other draw.
4. The probability of “success” (drawing a red marble) is the same for each trial:
This is assured by replacing each drawn marble back into the box before the next draw.
Since the scenario meets the criteria of a binomial experiment, the procedure will produce a
binomial distribution.
Solution: B. Procedure results in a binomial distribution
10. Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. Round to three decimal places.
n = 4, x = 3, p = 1/6 (Points : 4)
0.015
0.023
0.012
0.004
Solution: A. 0.015
11. Use the Poisson model to approximate the probability. Round your answer to four decimal places.
The probability that a call received by a certain switchboard will be a wrong number is 0.02. Use the Poisson distribution to approximate the probability that among 140 calls received by the switchboard, there are no wrong numbers. (Points : 4)
0.0608
0.1703
0.9392
0.8297
0.0669
The average number of wrong numbers received by the switchboard would be:
Then, using the Poisson model, the probability of zero wrong numbers would be:
Solution: A. 0.0608
12. Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05.
r = 0.843, n = 5 (Points : 4)
Critical values: r = ±0.878, significant linear correlation
Critical values: r = ±0.950, no significant linear correlation
Critical values: r = 0.950, significant linear correlation
Critical values: r = ±0.878, no significant linear correlation
From tables, with 5 – 2 = 3 degrees of freedom, the critical values are ±0.878.
Since the r value is between the two critical values, the null hypothesis is rejected and the conclusion is that there is a significant linear correlation.
Solution: A. Critical values: r = ±0.878, significant linear correlation
The paired data below consist of the test scores of 6 randomly selected students and the number of hours they studied for the test.
Hours - 5, 10, 4, 6, 10, 9
Scores - 64, 86, 69, 86, 59, 87
(Points : 4)
0.224
-0.678
0.678
-0.224
Complete the following table:
Hours (x) / Scores (y) / x^2 / xy / y^25 / 64 / 25 / 320 / 4096
10 / 86 / 100 / 860 / 7396
4 / 69 / 16 / 276 / 4761
6 / 86 / 36 / 516 / 7396
10 / 59 / 100 / 590 / 3481
9 / 87 / 81 / 783 / 7569
44 / 451 / 358 / 3345 / 34699
The correlation coefficient is then calculated from:
Solution: A. 0.224
14.If z is a standard normal variable, find the probability.
The probability that z is greater than -1.82 (Points : 4)
0.4656
0.9656
-0.0344
0.0344
Using a statistical calculator, the right tail area corresponding to z = -1.82 is 0.9656.
(Note: You can “cheat” a little on this problem. Since the z-score is negative, and you want the area to the right, you know that the answer has to be greater than 0.5. The only answer that would
work is B. 0.9656.)
Solution: B. 0.9656
15. Solve the problem.
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 40 different applicants are randomly selected, find the probability that their mean is above 215. (Points : 4)
0.0287
0.1179
0.4713
0.3821
Calculate the z-score:
Then, the probability that the mean is above 215 is equal to the probability that the z-score is greater than 1.8974:
Using a statistical calculator, the probability of a z-score being greater than 1.8974 is 0.0289
(Note: The “exact” value of 0.0287 comes from rounding the z-score off to 1.9.)
Solution: A. 0.0287
16. The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability.
The probability of no more than 75 defective CD's (Points : 4)
The area to the right of 75.5
The area to the left of 75
The area to the left of 74.5
The area to the left of 75.5
Another way of saying we want the probability of “no more than 75” defective CD’s, is to say that we want the probability of “≤ 75” defective CD’s.
To apply the continuity correction here, we need to include values up to 75.5.
As a result, the probability will be the area to the left of 75.5.
Solution: D. The area to the left of 75.5
17. Use the normal distribution to approximate the desired probability.
Find the probability that in 200 tosses of a fair die, we will obtain at most 30 fives. (Points : 4)
0.1871
0.4936
0.2946
0.3229
We want the probability of “at most 30 fives”, which is the same as saying we want the probability of “less than or equal to 30 fives”.
Applying the correction for continuity to this, we now want the probability of “less than 30.5 fives”.
The probability of tossing a five on a single roll is 1/6.
The z-score is calculated from:
with x = 30.5, n = 200, p = 1/6, and q = 1 – (1/6) = 5/6
The z-score is then:
Then, the probability of a z-score being less than -0.5376 is:
Again, the “exact” value of 0.2946 comes from rounding the z-score off to -0.54.
Solution: C. 0.2946
18. 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.
In a random sample of 184 college students, 97 had part-time jobs. Find the margin of error for the 95% confidence interval used to estimate the population proportion. (Points : 4)
0.0721
0.00266
0.126
0.0649
The margin of error is calculated from:
with for a 95% confidence interval, , ,
and n = 184.
Substituting those values gives:
Solution: A. 0.0721
19. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
n = 125, x = 72; 90% confidence (Points : 4)
0.507 < p < 0.645
0.503 < p < 0.649
0.506 < p < 0.646
0.502 < p < 0.650
For a 90% confidence interval, use
From the formula in Problem 18, with , , and n = 125, the
margin of error is:
The lower limit of the confidence interval is then:
LL =
The upper limit of the confidence interval is:
LL =
The confidence interval is then 0.503 < p < 0.649
Solution: B. 0.503 < p < 0.649
20. Use the given data to find the minimum sample size required to estimate the population proportion.
Margin of error: 0.015; confidence level: 96%; unknown.
6669
3667
4519
4670
Since and are unknown, use as a conservative estimate, as this will produce the largest sample size.
The sample size is calculated from:
The value of for a 96% confidence level is 2.05.
With , , and E = 0.015, the minimum sample size is:
Required sample size should always be rounded UP to the next whole number.
Solution: D. 4670
21. Solve the problem. Round the point estimate to the nearest thousandth.
50 people are selected randomly from a certain population and it is found that 13 people in the sample are over 6 feet tall. What is the point estimate of the proportion of people in the population who are over 6 feet tall? (Points : 4)
0.50
0.26
0.19
0.74
The point estimate is:
Solution: B. 0.26
22. Use the confidence level and sample data to find a confidence interval for estimating the population (mu). Round your answer to the same number of decimal places as the sample mean.
A random sample of 130 full-grown lobsters had a mean weight of 21 ounces and a standard deviation of 3.0 ounces. Construct a 98% confidence interval for the population mean mu. (Points : 4)
21 oz < mu < 23 oz
20 oz < mu < 22 oz
20 oz < mu < 23 oz
19 oz < mu < 21 oz
For a 98% confidence interval, use
The confidence interval is calculated from:
With , s = 3.0, and n = 130, the confidence interval is:
Rounded off to the same number of decimal places as the mean, the confidence interval is:
20 < µ < 22
Solution: B. 20 oz < mu < 22 oz
23. Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis.
= 0.05 for a left-tailed test. (Points : 4)
-1.96
±1.96
±1.645
-1.645
The critical z-value for a left-tailed test at = 0.05 is z = -1.645
Solution: D. -1.645
24. Find the value of the test statistic z using z =
The claim is that the proportion of accidental deaths of the elderly attributable to residential falls is more than 0.10, and the sample statistics include n = 800 deaths of the elderly with 15% of them attributable to residential falls. (Points : 4)
3.96
-3.96
4.71
-4.71
Calculate the test statistic using the formula:
with , , , and n = 800.
The value of the test statistic is then:
Solution: C. 4.71
25. Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).
The test statistic in a right-tailed test is z = 1.43. (Points : 4)
0.1528; fail to reject the null hypothesis
0.1528; reject the null hypothesis
0.0764; fail to reject the null hypothesis
0.0764; reject the null hypothesis
Using a statistical calculator, the p-value corresponding to a right-sided, one-tailed test with z = 1.43 is 0.0764.
Since this value is greater than the significance level, the decision would be to fail to reject the null hypothesis.
Solution: C. 0.0764, fail to reject the null hypothesis