The problem involves finding the probability that the sample mean of a certain number of observations (in this case, 55 families) falls within a certain range (17 to 18 pounds). This is a problem of sampling distributions, and we can solve it using the Central Limit Theorem and the concept of z-scores.

Here are the steps:

1. First, we need to calculate the standard error (SE) of the mean. The standard error is given by the formula SE = σ/√n, where σ is the standard deviation and n is the number of observations. In this case, σ = 2.5 pounds and n = 55. So, SE = 2.5/√55 = 0.3363 pounds.

2. Next, we calculate the z-scores for the lower and upper bounds of the range. The z-score is given by the formula z = (X - μ)/SE, where X is the value for which we want to find the z-score, μ is the mean, and SE is the standard error.

For the lower bound (X = 17 pounds), z = (17 - 17.2)/0.3363 = -0.5946.

For the upper bound (X = 18 pounds), z = (18 - 17.2)/0.3363 = 2.3770.

3. Finally, we find the probabilities corresponding to these z-scores from the standard normal distribution table (or using a calculator or software that can do this).

The probability for z = -0.5946 is 0.2764 (this is the probability that the sample mean is less than 17 pounds).

The probability for z = 2.3770 is 0.9914 (this is the probability that the sample mean is less than 18 pounds).

4. The probability that the sample mean is between 17 and 18 pounds is the difference between these two probabilities: 0.9914 - 0.2764 = 0.7150.

So, the probability that the mean of a sample of 55 families will be between 17 and 18 pounds is approximately 0.715 or 71.5%.

Question

The problem involves finding the probability that the sample mean of a certain number of observations (in this case, 55 families) falls within a certain range (17 to 18 pounds). This is a problem of sampling distributions, and we can solve it using the Central Limit Theorem and the concept of z-scores.

Here are the steps:

1. First, we need to calculate the standard error (SE) of the mean. The standard error is given by the formula SE = σ/√n, where σ is the standard deviation and n is the number of observations. In this case, σ = 2.5 pounds and n = 55. So, SE = 2.5/√55 = 0.3363 pounds.

2. Next, we calculate the z-scores for the lower and upper bounds of the range. The z-score is given by the formula z = (X - μ)/SE, where X is the value for which we want to find the z-score, μ is the mean, and SE is the standard error.

For the lower bound (X = 17 pounds), z = (17 - 17.2)/0.3363 = -0.5946.
   
   For the upper bound (X = 18 pounds), z = (18 - 17.2)/0.3363 = 2.3770.

3. Finally, we find the probabilities corresponding to these z-scores from the standard normal distribution table (or using a calculator or software that can do this).

The probability for z = -0.5946 is 0.2764 (this is the probability that the sample mean is less than 17 pounds).
   
   The probability for z = 2.3770 is 0.9914 (this is the probability that the sample mean is less than 18 pounds).

4. The probability that the sample mean is between 17 and 18 pounds is the difference between these two probabilities: 0.9914 - 0.2764 = 0.7150.

So, the probability that the mean of a sample of 55 families will be between 17 and 18 pounds is approximately 0.715 or 71.5%.

Knowee AI · Accepted Answer

The problem involves finding the probability that the sample mean of a certain number of observations (in this case, 55 families) falls within a certain range (17 to 18 pounds). This is a problem of sampling distributions, and we can solve it using the Central Limit Theorem and the concept of z-scores.

Here are the steps:

1. First, we need to calculate the standard error (SE) of the mean. The standard error is given by the formula SE = σ/√n, where σ is the standard deviation and n is the number of observations. In this case, σ = 2.5 pounds and n = 55. So, SE = 2.5/√55 = 0.3363 pounds.

2. Next, we calculate the z-scores for the lower and upper bounds of the range. The z-score is given by the formula z = (X - μ)/SE, where X is the value for which we want to find the z-score, μ is the mean, and SE is the standard error.

For the lower bound (X = 17 pounds), z = (17 - 17.2)/0.3363 = -0.5946.
   
   For the upper bound (X = 18 pounds), z = (18 - 17.2)/0.3363 = 2.3770.

3. Finally, we find the probabilities corresponding to these z-scores from the standard normal distribution table (or using a calculator or software that can do this).

The probability for z = -0.5946 is 0.2764 (this is the probability that the sample mean is less than 17 pounds).
   
   The probability for z = 2.3770 is 0.9914 (this is the probability that the sample mean is less than 18 pounds).

4. The probability that the sample mean is between 17 and 18 pounds is the difference between these two probabilities: 0.9914 - 0.2764 = 0.7150.

So, the probability that the mean of a sample of 55 families will be between 17 and 18 pounds is approximately 0.715 or 71.5%.

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