To solve this problem, we need to use the concept of the sampling distribution of the sample mean.

Step 1: Identify the parameters of the population distribution
The problem tells us that the heights of women in New Zealand are normally distributed with a mean (μ) of 165 cm and a standard deviation (σ) of 10 cm.

Step 2: Identify the parameters of the sampling distribution
The mean of the sampling distribution (μx̄) is equal to the population mean, which is 165 cm. The standard deviation of the sampling distribution (σx̄), also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size (n). In this case, n = 5, so σx̄ = σ/√n = 10/√5 ≈ 4.47 cm.

Step 3: Standardize the desired sample mean
We want to find the probability that the sample mean is greater than 162 cm. To do this, we first convert 162 cm to a z-score, which is a measure of how many standard errors the value is away from the mean of the sampling distribution. The z-score is calculated as follows: z = (x̄ - μx̄)/σx̄ = (162 - 165)/4.47 ≈ -0.67.

Step 4: Find the probability associated with the z-score
The z-score tells us that 162 cm is 0.67 standard errors below the mean. Because we want to find the probability that the sample mean is greater than 162 cm, we need to find the area to the right of this z-score on the standard normal distribution. Looking up -0.67 on a standard normal (Z) table or using a Z-table calculator, we find that the area to the left is approximately 0.2514. However, since we are interested in the area to the right, we subtract this value from 1 (because the total area under the curve is 1), giving us 1 - 0.2514 = 0.7486.

So, the probability that the sample mean height of five randomly selected women from New Zealand is greater than 162 cm is approximately 0.7486, or 74.86%.

Question

To solve this problem, we need to use the concept of the sampling distribution of the sample mean.

Step 1: Identify the parameters of the population distribution
The problem tells us that the heights of women in New Zealand are normally distributed with a mean (μ) of 165 cm and a standard deviation (σ) of 10 cm.

Step 2: Identify the parameters of the sampling distribution
The mean of the sampling distribution (μx̄) is equal to the population mean, which is 165 cm. The standard deviation of the sampling distribution (σx̄), also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size (n). In this case, n = 5, so σx̄ = σ/√n = 10/√5 ≈ 4.47 cm.

Step 3: Standardize the desired sample mean
We want to find the probability that the sample mean is greater than 162 cm. To do this, we first convert 162 cm to a z-score, which is a measure of how many standard errors the value is away from the mean of the sampling distribution. The z-score is calculated as follows: z = (x̄ - μx̄)/σx̄ = (162 - 165)/4.47 ≈ -0.67.

Step 4: Find the probability associated with the z-score
The z-score tells us that 162 cm is 0.67 standard errors below the mean. Because we want to find the probability that the sample mean is greater than 162 cm, we need to find the area to the right of this z-score on the standard normal distribution. Looking up -0.67 on a standard normal (Z) table or using a Z-table calculator, we find that the area to the left is approximately 0.2514. However, since we are interested in the area to the right, we subtract this value from 1 (because the total area under the curve is 1), giving us 1 - 0.2514 = 0.7486.

So, the probability that the sample mean height of five randomly selected women from New Zealand is greater than 162 cm is approximately 0.7486, or 74.86%.

Knowee AI · Accepted Answer

To solve this problem, we need to use the concept of the sampling distribution of the sample mean.

Step 1: Identify the parameters of the population distribution
The problem tells us that the heights of women in New Zealand are normally distributed with a mean (μ) of 165 cm and a standard deviation (σ) of 10 cm.

Step 2: Identify the parameters of the sampling distribution
The mean of the sampling distribution (μx̄) is equal to the population mean, which is 165 cm. The standard deviation of the sampling distribution (σx̄), also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size (n). In this case, n = 5, so σx̄ = σ/√n = 10/√5 ≈ 4.47 cm.

Step 3: Standardize the desired sample mean
We want to find the probability that the sample mean is greater than 162 cm. To do this, we first convert 162 cm to a z-score, which is a measure of how many standard errors the value is away from the mean of the sampling distribution. The z-score is calculated as follows: z = (x̄ - μx̄)/σx̄ = (162 - 165)/4.47 ≈ -0.67.

Step 4: Find the probability associated with the z-score
The z-score tells us that 162 cm is 0.67 standard errors below the mean. Because we want to find the probability that the sample mean is greater than 162 cm, we need to find the area to the right of this z-score on the standard normal distribution. Looking up -0.67 on a standard normal (Z) table or using a Z-table calculator, we find that the area to the left is approximately 0.2514. However, since we are interested in the area to the right, we subtract this value from 1 (because the total area under the curve is 1), giving us 1 - 0.2514 = 0.7486.

So, the probability that the sample mean height of five randomly selected women from New Zealand is greater than 162 cm is approximately 0.7486, or 74.86%.

he heights of women in New Zealand are normally distributed, with a mean of 165 centimetres and a standard deviation of 10 centimetres. A random sample of five women is selected. What is the probability that the sample mean is greater than 162 centimetres?

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