To solve this problem, we need to use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean.

Step 1: Identify the mean, standard deviation, and the value for which we need to find the probability. In this case, mean (μ) = $1200, standard deviation (σ) = $300.1, and the value (X) = $1600.

Step 2: Calculate the Z-score. The formula for Z-score is:

Z = (X - μ) / σ

Substituting the given values into the formula, we get:

Z = ($1600 - $1200) / $300.1 = 1.33 (rounded to two decimal places)

Step 3: Look up this Z-score in the Z-table to find the probability. The Z-table tells us the probability to the left of our Z-score. But we want the probability that the expenditure is more than $1600, so we want the area to the right of our Z-score.

The Z-table tells us that the probability to the left of Z = 1.33 is 0.9082. Therefore, the probability to the right is 1 - 0.9082 = 0.0918.

So, the probability that the weekly expenditure of a randomly selected household will be more than $1600 is 0.0918 or 9.18%.

Question

To solve this problem, we need to use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean.

Step 1: Identify the mean, standard deviation, and the value for which we need to find the probability. In this case, mean (μ) = $1200, standard deviation (σ) = $300.1, and the value (X) = $1600.

Step 2: Calculate the Z-score. The formula for Z-score is:

Z = (X - μ) / σ

Substituting the given values into the formula, we get:

Z = ($1600 - $1200) / $300.1 = 1.33 (rounded to two decimal places)

Step 3: Look up this Z-score in the Z-table to find the probability. The Z-table tells us the probability to the left of our Z-score. But we want the probability that the expenditure is more than $1600, so we want the area to the right of our Z-score.

The Z-table tells us that the probability to the left of Z = 1.33 is 0.9082. Therefore, the probability to the right is 1 - 0.9082 = 0.0918.

So, the probability that the weekly expenditure of a randomly selected household will be more than $1600 is 0.0918 or 9.18%.

Knowee AI · Accepted Answer

To solve this problem, we need to use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean.

Step 1: Identify the mean, standard deviation, and the value for which we need to find the probability. In this case, mean (μ) = $1200, standard deviation (σ) = $300.1, and the value (X) = $1600.

Step 2: Calculate the Z-score. The formula for Z-score is:

Z = (X - μ) / σ

Substituting the given values into the formula, we get:

Z = ($1600 - $1200) / $300.1 = 1.33 (rounded to two decimal places)

Step 3: Look up this Z-score in the Z-table to find the probability. The Z-table tells us the probability to the left of our Z-score. But we want the probability that the expenditure is more than $1600, so we want the area to the right of our Z-score.

The Z-table tells us that the probability to the left of Z = 1.33 is 0.9082. Therefore, the probability to the right is 1 - 0.9082 = 0.0918.

So, the probability that the weekly expenditure of a randomly selected household will be more than $1600 is 0.0918 or 9.18%.

Weekly household expenditure in a country is normally distributed with a mean $1200 and standard deviation of $300.1. What is the probability that the weekly expenditure of a randomly selected household will be more than $1600?

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