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The sampling distribution of the sample mean is developed by repeatedly taking s...
Sep 25, 2024
The sampling distribution of the sample mean is developed by repeatedly taking samples of size n and computing the sample means and reporting the resulting sample means in the form of a probability distribution.
Solution by Steps
step 1
The sampling distribution of the sample mean is formed by taking multiple samples of size n n from a population and calculating the mean of each sample
step 2
The means of these samples are then plotted to create a probability distribution, which is known as the sampling distribution of the sample mean
step 3
According to the Central Limit Theorem, as n n increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution, provided that the population has a finite mean and variance
Answer
The sampling distribution of the sample mean becomes approximately normal as the sample size increases.
Key Concept
The Central Limit Theorem is fundamental in understanding the behavior of sample means.
Explanation
It states that the distribution of sample means will tend to be normal as the sample size increases, which is crucial for making inferences about population parameters.
The sampling distribution of the sample mean is developed by repeatedly taking samples of size n and computing the sample means and reporting the resulting sample means in the form of a probability distribution. T or F
Solution by Steps
step 1
The statement is true because the sampling distribution of the sample mean is indeed developed by repeatedly taking samples of size n n from a population
step 2
Each sample mean is calculated and these means are then reported in the form of a probability distribution, which characterizes the sampling distribution of the sample mean
step 3
According to the Central Limit Theorem, as the sample size n n increases, the sampling distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution
Answer
True
Key Concept
The sampling distribution of the sample mean is formed by taking multiple samples and calculating their means.
Explanation
The statement accurately describes the process of developing the sampling distribution of the sample mean, confirming its truth.
Solution by Steps
step 1
The mean of the sampling distribution of xˉ \bar{x} is defined as μxˉ=μ \mu_{\bar{x}} = \mu , where μ \mu is the mean of the population from which the samples are drawn
step 2
This relationship holds true regardless of the sample size, as long as the samples are taken randomly and independently from the population
step 3
Therefore, the statement "The mean of the sampling distribution of xˉ \bar{x} is always equal to the mean of the sampled population" is true
Answer
True
Key Concept
The mean of the sampling distribution of the sample mean equals the population mean.
Explanation
This is a fundamental property of sampling distributions, ensuring that the expected value of the sample mean is the same as the population mean.
As the sample size increases, the standard deviation of the sampling distribution increases. T of F
Solution by Steps
step 1
The standard deviation of the sampling distribution of the sample mean, known as the standard error, is given by the formula σxˉ=σn \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} , where σ \sigma is the standard deviation of the population and n n is the sample size
step 2
As the sample size n n increases, the denominator n \sqrt{n} increases, which causes the standard error σxˉ \sigma_{\bar{x}} to decrease
step 3
Therefore, the statement "as the sample size increases, the standard deviation of the sampling distribution increases" is false
Answer
False
Key Concept
The standard deviation of the sampling distribution decreases as the sample size increases.
Explanation
This is due to the formula for the standard error, which shows that larger sample sizes lead to a smaller standard deviation of the sampling distribution.
For any sampled population, the population of all sample means is approximately normally distributed. T or F
Solution by Steps
step 1
According to the Central Limit Theorem, for any sampled population, as the sample size n n increases, the distribution of the sample means xˉ \bar{x} approaches a normal distribution, regardless of the shape of the population distribution
step 2
This means that even if the original population is not normally distributed, the distribution of the sample means will be approximately normally distributed if the sample size is sufficiently large (typically n≥30 n \geq 30 )
step 3
Therefore, the statement "the population of all sample means is approximately normally distributed" is true for large sample sizes
Answer
True
Key Concept
Central Limit Theorem
Explanation
The Central Limit Theorem states that the distribution of sample means will be approximately normal for large sample sizes, regardless of the population's distribution.
A random sample of size 36 is taken from a population with mean 50 and standard deviation 5. Find P( > 50.5).
Solution by Steps
step 1
We need to find P(\bar{x} > 50.5) for a sample size of n=36 n = 36 , population mean μ=50 \mu = 50 , and population standard deviation σ=5 \sigma = 5
step 2
First, we calculate the standard error (SE) of the sample mean using the formula: SE=σn=536=56≈0.8333 SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{36}} = \frac{5}{6} \approx 0.8333
step 3
Next, we standardize the value 50.5 50.5 to find the corresponding z-score using the formula: z=xˉ−μSE=50.5−500.8333≈0.6 z = \frac{\bar{x} - \mu}{SE} = \frac{50.5 - 50}{0.8333} \approx 0.6
step 4
Now, we find P(Z > 0.6) using the standard normal distribution table. The area to the left of z=0.6 z = 0.6 is approximately 0.7257 0.7257 , so P(Z > 0.6) = 1 - 0.7257 = 0.2743
Answer
The probability P(\bar{x} > 50.5) is approximately 0.2743 0.2743 .
Key Concept
The Central Limit Theorem allows us to use the normal distribution to approximate the sampling distribution of the sample mean.
Explanation
As the sample size increases, the distribution of the sample means approaches a normal distribution, enabling us to calculate probabilities using z-scores.
A random sample of size 36 is taken from a population with mean 50 and standard deviation 5. What is ?
Solution by Steps
step 1
To find σx \sigma_{x} , which represents the standard deviation of the sampling distribution of the sample mean, we use the formula: σx=σn \sigma_{x} = \frac{\sigma}{\sqrt{n}} , where σ \sigma is the population standard deviation and n n is the sample size
step 2
Given that σ=5 \sigma = 5 and n=36 n = 36 , we can substitute these values into the formula: σx=536=56≈0.8333 \sigma_{x} = \frac{5}{\sqrt{36}} = \frac{5}{6} \approx 0.8333
step 3
Therefore, the value of σx \sigma_{x} is approximately 0.8333 0.8333
Answer
D
Key Concept
Standard Error of the Mean
Explanation
The standard deviation of the sampling distribution of the sample mean is calculated using the population standard deviation divided by the square root of the sample size.
A random sample of size 36 is taken from a population with mean 50 and standard deviation 5. Find P( > 49).
Solution by Steps
step 1
We need to find P(\bar{x} > 49) for a sample size of n=36 n = 36 , population mean μ=50 \mu = 50 , and population standard deviation σ=5 \sigma = 5
step 2
First, we calculate the standard error (SE) of the sample mean using the formula: SE=σn=536=56≈0.8333 SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{36}} = \frac{5}{6} \approx 0.8333
step 3
Next, we standardize the value using the z-score formula: z=xˉ−μSE=49−500.8333≈−1.2001 z = \frac{\bar{x} - \mu}{SE} = \frac{49 - 50}{0.8333} \approx -1.2001
step 4
Now, we find P(\bar{x} > 49) = P(z > -1.2001) . Using the standard normal distribution table, P(z < -1.2001) \approx 0.1151 , thus P(z > -1.2001) = 1 - 0.1151 \approx 0.8849
Answer
The probability P(\bar{x} > 49) is approximately 0.8849 0.8849 .
Key Concept
The Central Limit Theorem allows us to use the normal distribution to find probabilities related to sample means.
Explanation
The answer shows how to calculate the probability of the sample mean being greater than a specific value using the standard error and z-scores.
The diameter of small Nerf balls manufactured overseas is expected to be approximately normally distributed with a mean of 5.2 inches and a standard deviation of .08 inches. Suppose a random sample of 20 balls is selected. Calculate the mean of the sampling distribution of the sample mean.
Solution by Steps
step 1
The mean of the sampling distribution of the sample mean xˉ \bar{x} is equal to the population mean μ \mu . Given that the population mean is μ=5.2 \mu = 5.2 inches, we have:
step 2
Therefore, the mean of the sampling distribution of the sample mean is μxˉ=μ=5.2 \mu_{\bar{x}} = \mu = 5.2 inches
Answer
The mean of the sampling distribution of the sample mean is 5.2 inches.
Key Concept
The mean of the sampling distribution of the sample mean equals the population mean.
Explanation
This means that regardless of the sample size, the average of the sample means will converge to the population mean.
The chief chemist for a major gasoline production company claims that the regular unleaded gasoline produced by the company contains on average 4 ounces of a certain ingredient. The chemist further states that the distribution of this ingredient per gallon of regular unleaded gasoline is normal and has a standard deviation of 1.2 ounces. What is the probability of finding an average of less than 3.85 ounces of this certain ingredient from 64 randomly inspected 1-gallon samples of regular unleaded gasoline? 答案选项组
Solution by Steps
step 1
We need to find the probability of the sample mean being less than 3.85 ounces. The sample mean Xˉ \bar{X} follows a normal distribution with mean μ=4 \mu = 4 ounces and standard deviation σ=1.2 \sigma = 1.2 ounces. The standard error (SE) of the sample mean for n=64 n = 64 is calculated as: SE=σn=1.264=1.28=0.15 SE = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{64}} = \frac{1.2}{8} = 0.15
step 2
Next, we standardize the sample mean using the z-score formula: z=Xˉ−μSE=3.85−40.15=−0.150.15=−1 z = \frac{\bar{X} - \mu}{SE} = \frac{3.85 - 4}{0.15} = \frac{-0.15}{0.15} = -1
step 3
Now, we will find the probability corresponding to the z-score of -1 using the standard normal distribution table. The probability P(Z < -1) is approximately 0.1587
Answer
The probability of finding an average of less than 3.85 ounces is approximately 0.1587.
Key Concept
The Central Limit Theorem allows us to use the normal distribution to find probabilities related to sample means.
Explanation
The calculated probability indicates that there is about a 15.87% chance of obtaining a sample mean of less than 3.85 ounces from the given population.
A random sample of size 36 is taken from a population with mean 50 and standard deviation 5. Find P( < 51.5). 答案选项组
Solution by Steps
step 1
We need to find the probability P(\bar{X} < 51.5) where Xˉ \bar{X} is the sample mean. Given the population mean μ=50 \mu = 50 and standard deviation σ=5 \sigma = 5 , we first calculate the standard error (SE) of the sample mean using the formula: SE=σn=536=56≈0.8333 SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{36}} = \frac{5}{6} \approx 0.8333
step 2
Next, we standardize the sample mean using the z-score formula: z=Xˉ−μSE=51.5−500.8333≈1.8 z = \frac{\bar{X} - \mu}{SE} = \frac{51.5 - 50}{0.8333} \approx 1.8
step 3
Now, we look up the z-score of 1.8 in the standard normal distribution table, which gives us the probability: P(Z < 1.8) \approx 0.9641
Answer
C
Key Concept
Sampling distribution of the sample mean and z-scores
Explanation
The sampling distribution of the sample mean is normally distributed for large samples, and we use z-scores to find probabilities associated with sample means.
P( < 25) if μ = 16 and = 4.
Solution by Steps
step 1
To find P(X < 25) , we first standardize the variable using the z-score formula: z=X−μσ z = \frac{X - \mu}{\sigma} , where μ=16 \mu = 16 and σ=4 \sigma = 4
step 2
Substituting the values into the z-score formula: z=25−164=94=2.25 z = \frac{25 - 16}{4} = \frac{9}{4} = 2.25
step 3
Now, we look up the z-score of 2.25 in the standard normal distribution table or use a calculator to find P(Z < 2.25) . This gives us approximately 0.9878 0.9878
Answer
The probability P(X < 25) is approximately 0.9878.
Key Concept
The z-score standardizes a value to find probabilities in a normal distribution.
Explanation
The z-score allows us to determine how many standard deviations a value is from the mean, enabling us to find the corresponding probability.
The diameter of small Nerf balls manufactured overseas is expected to be approximately normally distributed with a mean of 5.2 inches and a standard deviation of .08 inches. Suppose a random sample of 20 balls is selected. What percentage of sample means will be less than 5.14 inches?
Solution by Steps
step 1
First, we need to find the standard error of the sample mean. The standard error (SE) is calculated using the formula: SE=σn SE = \frac{\sigma}{\sqrt{n}} where σ=0.08 \sigma = 0.08 inches and n=20 n = 20
step 2
Calculating the standard error: SE=0.0820≈0.01789 SE = \frac{0.08}{\sqrt{20}} \approx 0.01789
step 3
Next, we will find the z-score for the sample mean of 5.14 inches using the formula: z=xˉ−μSE z = \frac{\bar{x} - \mu}{SE} where xˉ=5.14 \bar{x} = 5.14 inches and μ=5.2 \mu = 5.2 inches
step 4
Substituting the values into the z-score formula: z=5.14−5.20.01789≈−3.36 z = \frac{5.14 - 5.2}{0.01789} \approx -3.36
step 5
Now, we will find the percentage of sample means that are less than 5.14 inches by looking up the z-score in the standard normal distribution table. A z-score of -3.36 corresponds to a very small area
step 6
The area to the left of z=−3.36 z = -3.36 is approximately 0.0004, which means that about 0.04% of sample means will be less than 5.14 inches
Answer
Approximately 0.04% of sample means will be less than 5.14 inches.
Key Concept
The standard error and z-scores are crucial for determining probabilities related to sample means.
Explanation
The calculated z-score allows us to find the probability of obtaining a sample mean less than a specified value in a normal distribution.
The diameter of small Nerf balls manufactured overseas is expected to be approximately normally distributed with a mean of 5.2 inches and a standard deviation of .08 inches. Suppose a random sample of 20 balls is selected. Calculate the standard deviation of the sampling distribution of the sample mean.
Solution by Steps
step 1
To calculate the standard deviation of the sampling distribution of the sample mean, we use the formula: σxˉ=σn \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} where σ \sigma is the population standard deviation and n n is the sample size
step 2
Given that the population standard deviation σ=0.08 \sigma = 0.08 inches and the sample size n=20 n = 20 , we can substitute these values into the formula: σxˉ=0.0820 \sigma_{\bar{x}} = \frac{0.08}{\sqrt{20}}
step 3
Now, we calculate 20 \sqrt{20} : 20≈4.472 \sqrt{20} \approx 4.472
step 4
Substituting this value back into the formula gives us: σxˉ=0.084.472≈0.0179 \sigma_{\bar{x}} = \frac{0.08}{4.472} \approx 0.0179
Answer
The standard deviation of the sampling distribution of the sample mean is approximately 0.0179 inches.
Key Concept
The standard deviation of the sampling distribution of the sample mean quantifies the variability of sample means around the population mean.
Explanation
This value indicates how much the sample means are expected to vary from the true population mean, and it decreases as the sample size increases.
Packages of sugar bags for Sweeter Sugar Inc. have an average weight of 16 ounces and a standard deviation of .3 ounces. The weights of the sugar packages are normally distributed. What is the probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces?
Solution by Steps
step 1
We need to find the probability that the average weight of 9 randomly selected sugar packages exceeds 16.025 ounces. The population mean μ=16 \mu = 16 ounces and the population standard deviation σ=0.3 \sigma = 0.3 ounces
step 2
The standard deviation of the sampling distribution (standard error) for the sample mean is calculated using the formula: σxˉ=σn \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} where n=9 n = 9 . Thus, σxˉ=0.39=0.33=0.1 \sigma_{\bar{x}} = \frac{0.3}{\sqrt{9}} = \frac{0.3}{3} = 0.1
step 3
We will standardize the sample mean using the z-score formula: z=xˉ−μσxˉ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} For xˉ=16.025 \bar{x} = 16.025 , we have: z=16.025−160.1=0.0250.1=0.25 z = \frac{16.025 - 16}{0.1} = \frac{0.025}{0.1} = 0.25
step 4
Now, we will find the probability that the z-score is greater than 0.25. Using the standard normal distribution table, we find P(Z > 0.25) . The cumulative probability for z=0.25 z = 0.25 is approximately 0.5987, so: P(Z > 0.25) = 1 - 0.5987 = 0.4013
Answer
The probability that the average weight of 9 randomly selected packages exceeds 16.025 ounces is approximately 0.4013.
Key Concept
The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size increases, regardless of the population's distribution.
Explanation
In this problem, we applied the Central Limit Theorem to find the probability of the sample mean exceeding a certain value by calculating the z-score and using the standard normal distribution.
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