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Please show your work before checking the answer and explanations.
An Important Tip. Please draw a density curve and label all given in information on the curve before use the standard normal distribution table to find answers for any normal distribution related problems.
For samples of specified size from the population described, find the mean and standard deviation of the sample mean x̄..
Problem 1. .The mean and the standard deviation of the sampled population are, respectively, 115.3 and 36.5. n=81.
A) μx = 4.1; σx = 115.3
B) μx = 115.3; σx = 4.1
C) μx = 36.5; σx = 4.1
D) μx = 280.6; σx = 2.8
View Answer
Ans : B
Explanation:The mean is 115.3. The standard deviation is σ/√n, which is 36.5/√81. Therefore σx = 4.1 and the answer is B.
Problem 2. The National Weather Service keeps records of snowfall in mountain ranges. Records indicate that in a certain range, the annual snowfall has a mean of 110 inches and a standard deviation of 14 inches. Suppose the snowfalls are sampled during randomly picked years. For samples of size 49, determine the mean and standard deviation of x.
A) μx = 2 ; σx = 110
B) μx = 110 ; σx = 14
C) μx = 14 ; σx = 110
D) μx = 110 ; σx = 2
View Answer
Ans : D
Explanation:The mean is 110. The standard deviation is σ/√n, which is 14/√49. Therefore σx = 2 and the answer is D.
Problem 3. The National Weather Service keeps records of rainfall in valleys. Records indicate that in a certain valley, the annual rainfall has a mean of 95 inches and a standard deviation of 10 inches. Suppose the rainfalls are sampled during randomly picked years and x is the mean amount of rain in these years. For samples of size 25, determine the mean and standard deviation of x.
A) μx = 95; σx = 2
B) μx = 95; σx = 10
C) μx = 2 ; σx = 95
D) μx = 10; σx = 95
View Answer
Ans : A
Explanation:The mean is 95. The standard deviation is σ/√n, which is 10/√25. Therefore σx = 2 and the answer is A.
Problem 4. One truck from Lakeland Trucking, Inc. can carry a load of 4272.8 lbs. Records show that the weights of boxes that it carries have a mean of 83 lbs and a standard deviation of 14 lbs. For samples of size 49, find the mean and standard deviation of x.
A) μx = 2 ; σx = 83
B) μx = 83; σx = 14
C) μx = 83; σx = 2
D) μx = 14; σx = 83
View Answer
Ans : C
Explanation: The mean is 83. The standard deviation is σ/√n, which is 14/√49. Therefore σx = 2 and the answer is D.
Identify the distribution of the sample mean. In particular, state whether the distribution of x is normal or approximately normal and give its mean and standard deviation.
Problem 5. The weights of people in a certain population are normally distributed with a mean of 157 lbs and a standard deviation of 24 lbs. Determine the sampling distribution of the mean for samples of size 6.
A) Normal, mean = 157 lb, standard deviation = 24 lb
B) Approximately normal, mean = 157 lb, standard deviation = 4 lb
C) Approximately normal, mean = 157 lb, standard deviation = 9.8 lb
D) Normal, mean = 157 lb, standard deviation = 9.8 lb
View Answer
Ans : D
Explanation: Per the question, the population is normally distributed, so the mean is a normal mean=157. The standard deviation is &sigma/√n, which is 24/√6. Therefore σx = 9.8 and the answer is D.
Problem 6. The heights of people in a certain population are normally distributed with a mean of 68 inches and a standard deviation of 3.1 inches. Determine the sampling distribution of the mean for samples of size 44.
A) Normal, mean = 68 inches, standard deviation = 3.1 inches
B) Normal, mean = 68 inches, standard deviation = 0.07 inches
C) Normal, mean = 68 inches, standard deviation = 0.47 inches
D) Approximately normal, mean = 68 inches, standard deviation = 0.07 inches
View Answer
Ans : C
Explanation: Per the question, the population is normally distributed, so the mean is a normal mean=68. The standard deviation is σ/√n, which is 3.1/√44. Therefore σx = 0.47 and the answer is C.
Find the indicated probability or the percentage for the sampling distributions .
Problem 7. The distribution of weekly salaries at a large company is right skewed with a mean of $1000 and a standard deviation of $350. What is the probability that the mean of a random sample of weekly salaries of 50 employees will be at least $1050?
A) 1.0102
B) 0.8438
C) 0.1562
D) Cannot be determined, because the distribution of the population is not normal.
View Answer
Ans : C
Explanation: Standard Error is the standard deviation divided by the square root of n, which is 350/√50. Therefore the SE = 49.497. By doing z-score transformatio, z = (1050-1000)/49.497 = 1.0102. Use the normal table P(xbar > 1050) = P(Z>1.0102) = 1 - P(Z<1.0102) = 1 - 0.8438 = 0.1562.
Problem 8. Scores on an aptitude test are distributed with a mean of 220 and a standard deviation of 30. The shape of the distribution is unspecified. What is the probability that the mean of a random sample of 50 test scores will be at most 225 points?
A) 0.881
B) 0.119
C) 1.1785
D) Cannot be determined, because the distribution of the population is not known to be normal
View Answer
Ans : A
Explanation: Standard Error is the standard deviation divided by the square root of n, which is 30/√50. Therefore the SE = 4.243. To get the probability, we need to find z = (225-220)/4.243 = 1.1785. Use the z tables, the area to the left of 1.1785 is 0.8807.
Problem 9. The monthly expenditures on food by single adults in one city are normally distributed with a mean of $410 and a standard deviation of $70. What is the probability that the sampling error made in estimating the mean monthly expenditure of all single adults in that city by the mean of a random sample of 90 such adults will be at most $10?
A) -0.6776
B) 0.2490
C) 0.6776
D) 7.3786
View Answer
Ans : B
Explanation: Standard Error is the standard deviation divided by the square root of n, which is 70/√90. Therefore the SE = 7.37865. To get the probability, we need to find the z score of (405-410)/7.37865 = -0.6776. Use the z tables, we find P(Z < -0.6776) = 0.249.
Problem 10. The amount of coffee that a filling machine puts into an 8-ounce jar is normally distributed with a mean of 8.2 ounces and a standard deviation of 0.18 ounce. Determine the percentage of samples of size 16 that will have mean amounts of coffee within 8.1 ounce of the population mean of 8.3 ounces.
A) 98.68%
B) 42.46%
C) 71.23%
D) 97.36%
View Answer
Ans : D
Explanation: The lower limit is -0.1[.18/√16]=-2.22 and the limit is z 0.1[.18/√16] =2.22. Use your z table, R, or TI-84 calculator to find the area between the limits, which gives 97.36%.
Provide an appropiate response. .
Problem 11. The mean annual income for adult women in one city is $28520 and the standard deviation of the incomes is $5190 The distribution of incomes is skewed to the right. For samples of size 30, which of the following statements best describes the sampling distribution of the mean?
A) x is approximately normally distributed.
B) Nothing can be said about the distribution of x
C) The distribution of x is skewed to the right.
D) x is normally distributed.
View Answer
Ans : A
The sample size is 30, therefore via the central limit theorem x is approximately normally distributed.
Problem 12. Let x represent the number which shows up when a balanced die is rolled. Then x is a random variable with a uniform distribution. Let x denote the mean of the numbers obtained when the die is rolled 3 times. Which of the following statements concerning the sampling distribution of the mean, x, is true?
A) x is normally distributed.
B) x has a normal distribution.
C) x is approximately normally distributed.
D) None of the above statements are true
View Answer
Ans : D
There are only three roles, so none of the above statements (A-C) are correct.
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