Statistics Chapter 8 Homework
8.1 Complete parts (a) through (d) for the sampling distribution of the sample mean shown in the accompanying graph. Click the icon to view the graph. (a) What is the value of μx? The value of μx is ___ (b) What is the value of σx? The value of σx is ___ (c) If the sample size is n=16, what is likely true about the shape of the population? A. The shape of the population is approximately normal. B. The shape of the population is skewed left. C. The shape of the population is skewed right. D. The shape of the population cannot be determined. (d)If the sample size is n=16, what is the standard deviation of the population from which the sample was drawn? The standard deviation of the population from which the sample was drawn is _____.
(a) 500 (b) 10 (c)A. The shape of the population is approximately normal. (d) 40
8.1 The acceptable level for insect filth in a certain food item is 4 insect fragments (larvae, eggs, body parts, and so on) per 10 grams. A simple random sample of 40 ten-gram portions of the food item is obtained and results in a sample mean of x=4.9 insect fragments per ten-gram portion. Complete parts (a) through (c) below. Click here to view the standard normal distribution table (page 1).LOADING... Click here to view the standard normal distribution table (page 2).LOADING... . (a) Why is the sampling distribution of x approximately normal? A. The sampling distribution is approximately normal because the sample size is large enough. B. The sampling distribution is approximately normal because the population is normally distributed and the sample size is large enough. C. The sampling distribution is assumed to be approximately normal. D. The sampling distribution is approximately normal because the popluation is normally distributed. (b) What is the mean and standard deviation of the sampling distribution of x assuming μ=4 and σ=4? μx=___ (Round to three decimal places as needed.) σx=____ (Round to three decimal places as needed.) (c) What is the probability a simple random sample of 40 ten-gram portions of the food item results in a mean of at least 4.9 insect fragments? P(x≥4.9)=_____ (Round to four decimal places as needed.) Is this result unusual? A. This result is not unusual because its probability is large. B. This result is not unusual because its probability is small. C. This result is unusual because its probability is large. D. This result is unusual because its probability is small. What might we conclude? A. Since this result is not unusual, it is reasonable to conclude that the population mean is higher than 4. B. Since this result is unusual, it is reasonable to conclude that the population mean is higher than 4. C. Since this result is not unusual, it is not reasonable to conclude that the population mean is higher than 4. D. Since this result is unusual, it is not reasonable to conclude that the population mean is higher than 4.
(a) A. The sampling distribution is approximately normal because the sample size is large enough. Note: If a random variable X is normally distributed, the distribution of the sample mean, x, is normally distributed. If the sample size is large enough, n≥30, the sampling distribution is approximately normal regardless of the shape of the population. The sample size is greater than 30. (b) 4 ; 0.316 (c) 0.0022 D. This result is unusual because its probability is small. B. Since this result is unusual, it is reasonable to conclude that the population mean is higher than 4.
8.1 Consider a random variable X that is normally distributed. Complete parts (a) through (d) below. (This is a reading assessment question. Be certain of your answer because you only get one attempt on this question.) (a) If a random variable X is normally distributed, what will be the shape of the distribution of the sample mean? a. Skewed left b. Skewed right c. Normal d. Cannot be determined (b) If the mean of a random variable X is 55, what will be the mean of the sampling distribution of the sample mean? μx=____ (c) As the sample size n increases, what happens to the standard error of the mean? A. The standard error of the mean decreases. B. The standard error of the mean remains the same. C. The standard error of the mean increases (d) If the standard deviation of a random variable X is 25 and a random sample of size n=19 is obtained, what is the standard deviation of the sampling distribution of the sample mean? σx=____ (Type an exact answer, using radicals as needed.)
(a) c. Normal (b) 55 (c) A. The standard error of the mean decreases. Note: The correct answer is The standard error of the mean decreases because the standard error of the mean is defined as σx=σ / sqrt n, where σ is the standard deviation of the population and n is the sample size. As the sample size n increases, the standard error of the mean, σx, decreases. (d) 25/ sqrt 19
8.1 Without doing any computation, put the following in order from least to greatest, assuming the population is normally distributed with μ=300 and σ=25. (a) P(280≤x≤320) for a random sample of size n=20 (b) P(280≤x≤320) for a random sample of size n=40 (c) P(280≤x≤320)
(c)<(a)<(b)
8.1 Without doing any computation, put the following in order from least to greatest, assuming the population is normally distributed with μ=250 and σ=15. (a) P(240≤x≤260) for a random sample of size n=30 (b) P(240≤x≤260) for a random sample of size n=40 (c) P(240≤x≤260) ___ ____ ___
(c)<(a)<(b) Note: The standard deviation measures the spread in values. For a sample of size n, it is given by σ / sqrt n. If values are more spread out, then the probability of being near the mean is less likely.
8.1 A simple random sample of size n=49 is obtained from a population that is skewed left with μ=60 and σ=7. Does the population need to be normally distributed for the sampling distribution of x to be approximately normally distributed? Why? What is the sampling distribution of x? 1. Does the population need to be normally distributed for the sampling distribution of x to be approximately normally distributed? Why? A. Yes. The central limit theorem states that the sampling variability of nonnormal populations will increase as the sample size increases. B. Yes. The central limit theorem states that only for underlying populations that are normal is the shape of the sampling distribution of x normal, regardless of the sample size, n. C. No. The central limit theorem states that only if the shape of the underlying population is normal or uniform does the sampling distribution of x become approximately normal as the sample size, n, increases. D. No. The central limit theorem states that regardless of the shape of the underlying population, the sampling distribution of x becomes approximately normal as the sample size, n, increases. 2. What is the sampling distribution of x? Select the correct choice below and fill in the answer boxes within your choice. (Type integers or decimals rounded to three decimal places as needed.) A. The shape of the sampling distribution of x is unknown with μx=___ and σx=___. B. The sampling distribution of x is uniform with μx=__ and σx=___. C. The sampling distribution of x is approximately normal with μx=____ and σx=___. D. The sampling distribution of x is skewed left with μx=___ and σx=____.
1. D. No. The central limit theorem states that regardless of the shape of the underlying population, the sampling distribution of x becomes approximately normal as the sample size, n, increases. Note: If a random variable, X, is normally distributed, the distribution of the sample mean, x, is normally distributed. For nonnormal populations, review the conclusions of the central limit theorem. 2. C. The sampling distribution of x is approximately normal with μx=__60__ and σx=__1_.
8.1 For the three probability distributions shown, rank each distribution from lowest to highest in terms of the sample size required for the distribution of the sample mean to be approximately normally distributed. (a) A coordinate system labeled (a) has a horizontal axis labeled from 0 to 20 in increments of 20 and a vertical y-axis labeled from 0 to StartFraction 1 Over 20 EndFraction in increments of StartFraction 1 Over 20 EndFraction. A horizontal line extends from the y-axis at StartFraction 1 Over 20 EndFraction and a vertical line extends from the horizontal axis at 20. The area below the horizontal line and to the left of the vertical line is shaded. (b)μA normal curve labeled (b) a centered over a horizontal axis and centered on the mean. A vertical line segment extends from the horizontal axis to the curve at the mean. (c)A right tailed distribution is labeled (c). Choose the correct answer below. 1. (c), (a), (b) 2. (a), (b), (c) 3. (b), (c), (a) 4. (a), (c), (b) 5. (c), (b), (a) 6. (b), (a), (c)
1. (c), (a), (b) Note: The Central Limit Theorem states that, regardless of the shape of the underlying population, the sampling distribution of x becomes approximately normal as the sample size, n, increases. Two important conclusions to draw from this theorem are that 1) the shape of the distribution of the population from which the sample is drawn dictates the size of the sample required for the distribution of the sample mean to be normal, and 2) the more skewed the distribution of the population is, the larger the sample size needed to invoke the Central Limit Theorem. Use this information to determine the relative sample sizes necessary to normalize the given distributions.
According to a survey in a country, 26% of adults do not own a credit card. Suppose a simple random sample of 900 adults is obtained. Complete parts (a) through (d) below. Click here to view the standard normal distribution table (page 1).LOADING... Click here to view the standard normal distribution table (page 2).LOADING... (a) Describe the sampling distribution of p, the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of p below. A. Approximately normal because n≤0.05N and np(1−p)<10 B. Approximately normal because n≤0.05N and np(1−p)≥10 C. Not normal because n≤0.05N and np(1−p)<10 D. Not normal because n≤0.05N and np(1−p)≥10 Determine the mean of the sampling distribution of p. μp=___ (Round to two decimal places as needed.) Determine the standard deviation of the sampling distribution of p. σp=___ (Round to three decimal places as needed.) (b) What is the probability that in a random sample of 900 adults, more than 29% do not own a credit card? The probability is ____. (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 900 adults were obtained, one would expect ___ to result in more than 29% not owning a credit card. (Round to the nearest integer as needed.) (c) What is the probability that in a random sample of 900 adults, between 24% and 29% do not own a credit card? The probability is ___. (Round to four decimal places as needed.) Interpret this probability. If 100 different random samples of 900 adults were obtained, one would expect___ to result in between 24% and 29% not owning a credit card. (Round to the nearest integer as needed.) (d) Would it be unusual for a random sample of 900 adults to result in 216 or fewer who do not own a credit card? Why? Select the correct choice below and fill in the answer box to complete your choice. (Round to four decimal places as needed.) A. The result is unusual because the probability that p is less than or equal to the sample proportion is ____, which is greater than 5%. B. The result is unusual because the probability that p is less than or equal to the sample proportion is ___, which is less than 5%. C. The result is not unusual because the probability that p is less than or equal to the sample proportion is ____, which is greater than 5%. D. The result is not unusual because the probability that p is less than or equal to the sample proportion is ____, which is less than 5%.
B. Approximately normal because n≤0.05N and np(1−p)≥10 0.26 0.015 0.0228 2 0.8860 89 c;0.0918
8.2 Describe the sampling distribution of p. Assume the size of the population is 25,000. n=500, p=0.737 Describe the shape of the sampling distribution of p. Choose the correct answer below. A. The shape of the sampling distribution of p is approximately normal because n≤0.05N and np(1−p)<10. B. The shape of the sampling distribution of p is approximately normal because n≤0.05N and np(1−p)≥10. Your answer is correct. C. The shape of the sampling distribution of p is not normal because n≤0.05N and np(1−p)≥10. D. The shape of the sampling distribution of p is not normal because n≤0.05N and np(1−p)<10. Determine the mean of the sampling distribution of p. μp=____ (Round to three decimal places as needed.) Determine the standard deviation of the sampling distribution of p. σp=_____ (Round to three decimal places as needed.)
B. The shape of the sampling distribution of p is approximately normal because n≤0.05N and np(1−p)≥10. 0.737 0.02
8.2 Suppose a simple random sample of size n=200 is obtained from a population whose size is N=30,000 and whose population proportion with a specified characteristic is p=0.8. Click here to view the standard normal distribution table (page 1).LOADING... Click here to view the standard normal distribution table (page 2).LOADING... (a) Describe the sampling distribution of p. Choose the phrase that best describes the shape of the sampling distribution below. A. Approximately normal because n≤0.05N and np(1−p)<10. B. Not normal because n≤0.05N and np(1−p)≥10. C. Approximately normal because n≤0.05N and np(1−p)≥10. Your answer is correct. D. Not normal because n≤0.05N and np(1−p)<10. Determine the mean of the sampling distribution of p. μp=___(Round to one decimal place as needed.) Determine the standard deviation of the sampling distribution of p. σp=____ (Round to six decimal places as needed.) (b) What is the probability of obtaining x=170 or more individuals with the characteristic? That is, what is P(p≥0.85)? P(p≥0.85)=_____ (Round to four decimal places as needed.) (c) What is the probability of obtaining x=154 or fewer individuals with the characteristic? That is, what is P(p≤0.77)? P(p≤0.77)=_____ (Round to four decimal places as needed.)
C. Approximately normal because n≤0.05N and np(1−p)≥10. 0.8 0.028284 0.0384 0.1446
8.2 According to a study, the proportion of people who are satisfied with the way things are going in their lives is 0.82. Suppose that a random sample of 100 people is obtained. Complete parts (a) through (e) below. Click here to view the standard normal distribution table (page 1).LOADING... Click here to view the standard normal distribution table (page 2).LOADING... (a) Suppose the random sample of 100 people is asked, "Are you satisfied with the way things are going in your life?" Is the response to this question qualitative or quantitative? Explain. A. The response is quantitative because the responses can be classified based on the characteristic of being satisfied or not. B. The response is qualitative because the number of people satisfied can be counted. C. The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not. D. The response is quantitative because the number of people satisfied can be counted. (b) Explain why the sample proportion, p, is a random variable. What is the source of the variability? A. The sample proportion p is a random variable because the value of p varies from sample to sample. The variability is due to the fact that people may not be responding to the question truthfully. B. The sample proportion p is a random variable because the value of p varies from sample to sample. The variability is due to the fact that different people feel differently regarding their satisfaction. C. The sample proportion p is a random variable because the value of p represents a random person included in the sample. The variability is due to the fact that people may not be responding to the question truthfully. D. The sample proportion p is a random variable because the value of p represents a random person included in the sample. The variability is due to the fact that different people feel differently regarding their satisfaction. (c) Describe the sampling distribution of p, the proportion of people who are satisfied with the way things are going in their life. Be sure to verify the model requirements. Since the sample size is ___ 5% of the population size and np(1−p)=_____≥10, the distribution of p is______ with μp=____and σp=___ (Round to three decimal places as needed.) (d) In the sample obtained in part (a), what is the probability that the proportion who are satisfied with the way things are going in their life exceeds 0.84? The probability that proportion who are satisfied with the way things are going in their life exceeds 0.84 is ____. (Round to four decimal places as needed. (e) Using the distribution from part (c), would it be unusual for a survey of 100 people to reveal that 74 or fewer people in the sample are satisfied with their lives? The probability that 74 or fewer people in the sample are satisfied is ____, which ____ unusual because this probability ____ less than ____. (Round to four decimal places as needed.)
C. The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not. B. The sample proportion p is a random variable because the value of p varies from sample to sample. The variability is due to the fact that different people feel differently regarding their satisfaction. Note: The sample proportion, p, is a statistic that estimates the population proportion, p. Review the given situation and determine why the sample proportion is a random variable, as well as what causes it to vary. Since the sample size is ___no more_ than 5% of the population size and np(1−p)=_+14.76__≥10, the distribution of p is __approximately normal__ with μp=__0.82__ and σp=__0.038.___ Note:Let p be the proportion of people who are satisfied with the way things are going in their lives, and let n be the size of the sample. For a simple random sample of size n with independent sample values and a population proportion p, the shape of the sampling distribution is approximately normal provided np(1−p)≥10. The mean of the sampling distribution is μp=p. The standard deviation of the sampling distribution of p is given by the following formula. Remember to round to three decimal places. 0.3015 The probability that 74 or fewer people in the sample are satisfied is __0.0188__, which _is_ unusual because this probability __is__ less than _5_ %.
8.2 True or False: The population proportion and sample proportion always have the same value. Choose the correct answer below. False True
False Note: The population proportion and sample proportion do not always have the same value.
8.2 Describe the sampling distribution of p. Assume the size of the population is 15,000. n=300, p=0.9
For a simple random sample of size n such that n≤0.05N, where N is the size of the population, the shape of the sampling distribution of p is approximately normal provided np(1−p)≥10. Determine 5% of the population size. 0.05(15,000)=750 Evaluate np(1−p). np(1−p) = (300)(0.9)(1−0.9) = (300)(0.9)(0.1) = 27 The shape of the sampling distribution of p is approximately normal because np(1−p)≥10 and the sample is less than 5% of the population. The mean of the sampling distribution of p is the same as the population proportion. μp=0.9 Determine the standard deviation of the sampling distribution of p, rounding to three decimal places. σp = sqrt p(1−p)n = sqrt 0.9(1−0.9)300 = sqrt 0.0003 = 0.017 Therefore, the sampling distribution of p is approximately normal with mean μp=0.9 and standard deviation σp=0.017.
8.1 A simple random sample of size n=50 is obtained from a population that is skewed left with μ=50 and σ=11. Does the population need to be normally distributed for the sampling distribution of x to be approximately normally distributed? Why? What is the sampling distribution of x?
If a random variable, X, is normally distributed, the distribution of the sample mean, x, is normally distributed. Further, the central limit theorem states that the sampling distribution of the sample mean, x, becomes approximately normal as the sample size, n, increases, regardless of the shape of the underlying population. Use this to determine if the population needs to be normally distributed for the sampling distribution of x to be approximately normally distributed. To determine the sampling distribution of x, first determine its shape. The shape of the sampling distribution of x is approximately normal. Suppose that a simple random sample of size n is drawn from a population with mean μ and standard deviation σ. The sampling distribution of x has a mean of μx=μ and a standard deviation given by the formula below. σx=σ / sqrt n Determine the mean of the sampling distribution of x. μx=50 Calculate σx, the standard deviation of the sampling distribution of x. Begin by substituting values for σ and n into the given formula for σx. σx =σ / sqrt n = 11 / sqrt 50 Simplify to find the standard deviation of the sampling distribution of x, rounding to three decimal places. σx =11 / sqrt 50 = 1.556 Therefore, the sampling distribution of x is normal or approximately normal with mean μx=50 and standard deviation σx=1.556.
8.1 Complete the sentence below. The standard deviation of the sampling distribution of x, denoted σx, is called the _____ _____ of the _____.
Standard; error; mean Note: The sampling distribution of the sample mean x is the probability distribution of all possible values of the random variable x computed from a sample of size n from a population with mean μ and standard deviation σ. Suppose that a simple random sample of size n is drawn from a large population with mean μ and standard deviation σ. The sampling distribution of x will have mean μx=μ and standard deviation σx=σ / sqrt n. The standard deviation of the sampling distribution of x, σx, is called the standard error of the mean.
8.2 Complete the sentence below. The _____ _____, denoted p, is given by the formula p=_____, where x is the number of individuals with a specified characteristic in a sample of n individuals.
The __sample proportion__ ,denoted p,is given by the formula ___p=x/n ___where x is the number of individuals with a specified characteristic in a sample of n individuals. Note: Recall that the population mean is denoted μ, the sample mean is denoted x, the population proportion is denoted p, the population variance is denoted σ2, the sample variance is denoted s2, and the population and sample standard deviations are denoted σ and s, respectively. Note that the sample proportion is a statistic that estimates the population proportion, p. Also, a population proportion is the number of individuals in a population with a specified characteristic divided by the number of individuals in the population. Suppose that a random sample of size n is obtained from a population in which each individual either does or does not have a certain characteristic. The sample proportion, denoted p, is given by p=x/n, where x is the number of individuals with a specified characteristic in a sample of n individuals. The sample proportion, p, is a statistic that estimates the population proportion, p.
8.1 Is the statement below true or false? The distribution of the sample mean, x, will be normally distributed if the sample is obtained from a population that is normally distributed, regardless of the sample size. Choose the correct answer below. True False
True Note: The statement is true. The sampling distribution of the sample mean x is the probability distribution of all possible values of the random variable x computed from a sample of size n from a population with mean μ and standard deviation σ. Suppose that a simple random sample of size n is drawn from a large population with mean μ and standard deviation σ. The sampling distribution of x will have mean μx=μ and standard deviation σx=σ / sqrt n. If a random variable X is normally distributed, the distribution of the sample mean, x, is normally distributed.
8.1 The acceptable level for insect filth in a certain food item is 5 insect fragments (larvae, eggs, body parts, and so on) per 10 grams. A simple random sample of 40 ten-gram portions of the food item is obtained and results in a sample mean of x=5.5 insect fragments per ten-gram portion. Complete parts (a) through (c) below. Click here to view the standard normal distribution table (page 1).LOADING... Click here to view the standard normal distribution table (page 2).LOADING... (a) Why is the sampling distribution of x approximately normal? A. The sampling distribution is assumed to be approximately normal. B. The sampling distribution is approximately normal because the sample size is large enough. Your answer is correct. C. The sampling distribution is approximately normal because the popluation is normally distributed. D. The sampling distribution is approximately normal because the population is normally distributed and the sample size is large enough. (b) What is the mean and standard deviation of the sampling distribution of x assuming μ=5 and σ=5? μx=5 (Round to three decimal places as needed.) σx=0.354 (Round to three decimal places as needed.) (c) What is the probability a simple random sample of 40 ten-gram portions of the food item results in a mean of at least 5.5 insect fragments? P(x≥5.5)=0.0793 (Round to four decimal places as needed.) Is this result unusual? A. This result is unusual because its probability is large. B. This result is not unusual because its probability is small. C. This result is not unusual because its probability is large. Your answer is correct. D. This result is unusual because its probability is small. What might we conclude? A. Since this result is not unusual, it is reasonable to conclude that the population mean is higher than 5. B. Since this result is not unusual, it is not reasonable to conclude that the population mean is higher than 5. Your answer is correct. C. Since this result is unusual, it is reasonable to conclude that the population mean is higher than 5. D. Since this result is unusual, it is not reasonable to conclude that the population mean is higher than 5.
a) B. The sampling distribution is approximately normal because the sample size is large enough. b) μx=5 σx=0.354 c) P(x≥5.5)=0.0793 C. This result is not unusual because its probability is large. B. Since this result is not unusual, it is not reasonable to conclude that the population mean is higher than 5. Note:If a result is unusual, it is reasonable to believe that the assumed mean is not correct. Otherwise, it is not reasonable to believe that the assumed mean is not correct.
8.1 Suppose Jack and Diane are each attempting to use a simulation to describe the sampling distribution from a population that is skewed right with mean 50 and standard deviation 15. Jack obtains 1000 random samples of size n=5 from the population, finds the mean of the means, and determines the standard deviation of the means. Diane does the same simulation, but obtains 1000 random samples of size n=30 from the population. Complete parts (a) through (c). (a) Describe the shape you expect for Jack's distribution of sample means. Describe the shape you expect for Diane's distribution of sample means. Choose the correct answer below. A. Jack's distribution and Diane's distribution are expected to be approximately normal. However, Jack's will have the greater standard deviation of the sample mean. B. Jack's distribution and Diane's distribution are expected to be approximately normal. However, Jack's will have the smaller standard deviation of the sample mean. C. Jack's distribution is expected to be skewed right, but less skewed than the original distribution. Diane's distribution is expected to be approximately normal. D. Diane's distribution is expected to be skewed right, but less skewed than the original distribution. Jack's distribution is expected to be approximately normal. (b) What do you expect the mean of Diane's distribution to be? What do you expect the mean of Jack's distribution to be? Diane's distribution is expected to have a mean of ___. Jack's distribution is expected to have a mean of ___. (Type integers or decimals rounded to two decimal places as needed.) (c) What do you expect the standard deviation of Diane's distribution to be? What do you expect the standard deviation of Jack's distribution to be? Diane's distribution is expected to have a standard deviation of _____. Jack's distribution is expected to have a standard deviation of____. (Type integers or decimals rounded to two decimal places as needed.)
a)D. Diane's distribution is expected to be skewed right, but less skewed than the original distribution. Jack's distribution is expected to be approximately normal. Note: Since Diane's sample size is small, the shape of the sampling distribution of x will still be affected by the shape of the population distribution. Since Jack's sample size is large, the shape of the sampling distribution of x will be nearly normal. The Central Limit Theorem states that regardless of the shape of the underlying population, the sample distribution of x becomes approximately normal as the sample size, n, increases. b) 60; 60 c) 6.71;2.74
8.1 Suppose a simple random sample of size n=12 is obtained from a population with μ=61 and σ=17. (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities regarding the sample mean? Assuming the normal model can be used, describe the sampling distribution x. (b) Assuming the normal model can be used, determine P(x<64.8). (c) Assuming the normal model can be used, determine P(x≥63.4). Click here to view the standard normal distribution table (page 1).LOADING... Click here to view the standard normal distribution table (page 2).LOADING... (a) What must be true regarding the distribution of the population? A. The population must be normally distributed and the sample size must be large. B. Since the sample size is large enough, the population distribution does not need to be normal. C. The population must be normally distributed. D. There are no requirements on the shape of the distribution of the population. Assuming the normal model can be used, describe the sampling distribution x. A. Normal, with μx=61 and σx=1217 B. Normal, with μx=61 and σx=1712 C. Normal, with μx=61 and σx=17 (b) P(x<64.8)=______ (Round to four decimal places as needed.) (c) P(x≥63.4)=____ (Round to four decimal places as needed.)
a. C. The population must be normally distributed. B. Normal, with μx=61 and σx=17 / qrt12 b)0.7806 c)0.3124
8.1 Fill in the blanks to correctly complete the sentence below. Suppose a simple random sample of size n is drawn from a large population with mean μ and standard deviation σ. The sampling distribution of x has mean μx=______ and standard deviation σx=______.
μ ; σ / sqrt (n) Note: For a simple random sample of size n that is drawn from a large population with mean μ and standard deviation σ, the sampling distribution of x has mean μx=μ and standard deviation σx = σ / sqrt (n). Notice that as n increases, σx decreases. The sample mean matches the population mean, but the standard error of the mean depends on the sample size. Notice that as n increases, σx decreases.
8.2 According to a survey in a country, 42% of adults do not own a credit card. Suppose a simple random sample of 800 adults is obtained. Complete parts (a) through (d) below. Click here to view the standard normal distribution table (page 1).LOADING... Click here to view the standard normal distribution table (page 2).LOADING...
(a) Describe the sampling distribution of p, the sample proportion of adults who do not own a credit card. For a simple random sample size n such that n≤0.05N, the shape of the sampling distribution of p is approximately normal provided np(1−p)≥10. The sample size 800 is 5% of 16,000. It is safe to assume there are more than 16,000 adults in the country so the sample size is less than 5% of the population size. Evaluate np(1−p). np(1−p) = (800)(0.42)(1−0.42) = 194.88 The shape of the sampling distribution of p is approximately normal. The mean of the sampling distribution of p is the same as the population proportion. μp=0.42 Determine the standard deviation of the sampling distribution of p, rounding to three decimal places. σp = p(1−p)n = 0.42(1−0.42)800 = 0.017 (b) What is the probability that in a random sample of 800 adults, more than 44% do not own a credit card? While either technology or a standard normal distribution table can be used to find the probability, for this problem, use the table. The normal curve with p=0.44 is shown to the right. The area to the right of p=0.44 corresponds to P(p>0.44). A normal curve is over a horizontal axis labeled p hat. Vertical line segments extend from the horizontal axis to the curve at the mean, and 0.44, where 0.39 is to the right of the mean Convert p=0.44 to a standard normal random variable Z, rounding to two decimal places. z = p−μp / σp = 0.44−0.42 / 0.017 = 1.18 The table value for z=1.18, or P(p<0.44), is 0.8810. The area to the right of z=1.18 corresponds to P(p>0.44). P(p>0.44)=1−0.8810=0.1190 So the probability that in a random sample of 800 adults, more than 44% do not own a credit card is 0.1190. To interpret the probability, determine the number of times out of 100 this outcome would be expected. Interpret the probability that in a random sample of 800 adults, more than 44% do not own a credit card, rounding to the nearest integer. If 100 different random samples of 800 adults were obtained, one would expect 12 to result in more than 44% not owning a credit card. (c) What is the probability that in a random sample of 800 adults, between 39% and 44% do not own a credit card? While either technology or a standard normal distribution table can be used to find the probability, for this problem, use the table. To find the probability that the percent of the random sample of 800 adults that do not own a credit card is between 39% and 44%, subtract the probability that the percent is lower than 39% from the probability that the percent is lower than 44%. In part (b), the probability that the percent is lower than 44% was found to be 0.8810. The normal curve with p=0.39 is shown to the right. The area to the left of p=0.39 corresponds to P(p<0.39). A normal curve is over a horizontal axis labeled p hat. Vertical line segments extend from the horizontal axis to the curve at the mean, and 0.39, where 0.39 is to the left of the mean. Convert p=0.39 to a standard normal random variable Z, rounding to two decimal places. z = p−μp / σp = 0.39−0.42 / 0.017 = −1.76 The area to the left of z=−1.76 corresponds to P(p<0.39). Using the standard normal distribution table, this probability is P(p<0.39)=0.0392. Lastly, subtract P(p<0.39) from P(p<0.44) to find P(0.39<p<0.44). 0.8810−0.0392=0.8418 So the probability that in a random sample of 800 adults, between 39% and 44% do not own a credit card is 0.8418. To interpret the probability, determine the number of times out of 100 this outcome would be expected. Interpret the probability that in a random sample of 800 adults, between 39% and 44% do not own a credit card, rounding to the nearest integer. If 100 different random samples of 800 adults were obtained, one would expect 84 to result in between 39% and 44% not owning a credit card. (d) Would it be unusual for a random sample of 800 adults to result in 312 or fewer who do not own a credit card? Why? To determine the probability that 312 or more adults in a sample of 800 do not own a credit card, first find the sample proportion. p=312 / 800=0.39 Recall that, in part (c), P(p<0.39) was found to be 0.0392. Note that an unusual event is an event that has a low probability of occurring. Typically, an event with a probability less than 0.05 (or 5%) is considered unusual. Use this information to determine if the result is unusual.
8.1 Complete parts (a) through (d) for the sampling distribution of the sample mean shown in the accompanying graph. Click the icon to view the graph.
(a) Determine the value of μx. To find the value of μx, determine the center of the distribution of the sample means. μx=200 (b) Determine the value of σx. Note that the left and right points shown on the graph are the inflection points. Recall that the inflection points are the points on the curve where the curvature of the graph changes. For a normal curve, the inflection points are the points at x=μ−σ and x=μ+σ. To the left of x=μ−σ and to the right of x=μ+σ, the curve is drawn upward. In between, the curve is drawn downward. From the problem statement graph, x=225 for the right point. From part (a), μx=200. Substitute these values into the formula for the right point and find the standard deviation of the sampling distribution of x, σx. x= μx+σx 225=200+σx σx=225−200 σx=25 (c) If the sample size is n=25, determine what must be true about the shape of the population. The distribution of the sample mean is normally distributed when the random variable X is normally distributed, or when the sample size is large (at least 30). Notice that in this case the distribution of the sample mean is normal and the sample size is small. The sampling distribution of the mean is only normal when the population is normal or the sample size is large. Thus, the shape of the population must be approximately normal because the sample size is not large. (d) If the sample size is n=25, determine the standard deviation of the population from which the sample was drawn. Recall that for a simple random sample of size n drawn from a large population with mean μ and standard deviation σ, the sampling distribution of x will have mean μx=μ and standard deviation σx=σn. The standard deviation of the sampling distribution of x, σx, is called the standard error of the mean. Use this information to find the value of the population standard deviation σ. From part b, σx=25. σx=σ / sqrt n 25=σ/ sqrt 25 σ=25• sqrt25 σ=125 Thus, the standard deviation of the population from which the sample was drawn is 125.
8.1 Suppose a simple random sample of size n=17 is obtained from a population with μ=71 and σ=12. (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities regarding the sample mean? Assuming the normal model can be used, describe the sampling distribution x. (b) Assuming the normal model can be used, determine P(x<73.7). (c) Assuming the normal model can be used, determine P(x≥72.8). Click here to view the standard normal distribution table (page 1).LOADING... Click here to view the standard normal distribution table (page 2).LOADING...
(a) What must be true regarding the distribution of the population? If a random variable X is normally distributed, the distribution of the sample mean, x, is normally distributed. If the sample size is large enough, n≥30, the sampling distribution is approximately normal regardless of the shape of the population. The sample size must be greater than or equal to 30 or the population must be normally distributed in order to use the normal model to compute probabilities regarding the sample mean. Assuming the normal model can be used, the sampling distribution of x has mean μx=μ and standard deviation σx=σn. Identify μx. μx=71 Calculate σx. Substitute the values of σ and n and simplify. σx =σ / sqrt n =12 / sqrt 17 Substitute. ≈2.91 Simplify. (b) To determine P(x<73.7), first convert x=73.7 to a Z-score. Z= (x−μx) / σx =73.7−71 / 2.91 Substitute. ≈0.93 Simplify. The normal curve with μx=71 and σx=2.91 is shown on the right. The area under the curve to the left of 73.7 corresponds with P(x<73.7). Use a standard normal distribution table or similar means to determine the area under the standard normal curve to the left of Z=0.93. P(x<73.7)≈0.8238 (c) To determine P(x≥72.8), first convert x=72.8 to a Z-score. Z =(x−μx) / σx =72.8−71 / 2.91 Substitute. ≈ 0.62 Simplify. The normal curve with μx=71 and σx=2.91 is shown on the right. The area under the curve to the right of 72.8 corresponds with P(x≥72.8). Use a standard normal distribution table or similar means to determine the area under the standard normal curve to the right of Z=0.62. P(x≥72.8)≈1−0.7324 =0.2676
