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

Step 1: Calculate the Z-score for 0.23 inches. The formula for the Z-score is:

Z = (X - μ) / σ

where:
X is the value we are interested in (0.23 inches in this case),
μ is the mean (0.22 inches), and
σ is the standard deviation (0.05 inches).

So, Z = (0.23 - 0.22) / 0.05 = 0.2

Step 2: Look up this Z-score in the Z-table to find the proportion of data below this score. A Z-score of 0.2 corresponds to a proportion of approximately 0.5793. This means that approximately 57.93% of woodlice have antenna lengths less than 0.23 inches.

Step 3: Since we want to find the proportion of woodlice with antenna lengths that are at least 0.23 inches, we subtract the proportion we found from 1 (since the total proportion is always 1).

So, 1 - 0.5793 = 0.4207

Therefore, approximately 42.07% of woodlice have antenna lengths that are at least 0.23 inches. Rounded to four decimal places, this is 0.4207.

Question

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

Step 1: Calculate the Z-score for 0.23 inches. The formula for the Z-score is:

Z = (X - μ) / σ

where:
X is the value we are interested in (0.23 inches in this case),
μ is the mean (0.22 inches), and
σ is the standard deviation (0.05 inches).

So, Z = (0.23 - 0.22) / 0.05 = 0.2

Step 2: Look up this Z-score in the Z-table to find the proportion of data below this score. A Z-score of 0.2 corresponds to a proportion of approximately 0.5793. This means that approximately 57.93% of woodlice have antenna lengths less than 0.23 inches.

Step 3: Since we want to find the proportion of woodlice with antenna lengths that are at least 0.23 inches, we subtract the proportion we found from 1 (since the total proportion is always 1).

So, 1 - 0.5793 = 0.4207

Therefore, approximately 42.07% of woodlice have antenna lengths that are at least 0.23 inches. Rounded to four decimal places, this is 0.4207.

Knowee AI · Accepted Answer

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

Step 1: Calculate the Z-score for 0.23 inches. The formula for the Z-score is:

Z = (X - μ) / σ

where:
X is the value we are interested in (0.23 inches in this case),
μ is the mean (0.22 inches), and
σ is the standard deviation (0.05 inches).

So, Z = (0.23 - 0.22) / 0.05 = 0.2

Step 2: Look up this Z-score in the Z-table to find the proportion of data below this score. A Z-score of 0.2 corresponds to a proportion of approximately 0.5793. This means that approximately 57.93% of woodlice have antenna lengths less than 0.23 inches.

Step 3: Since we want to find the proportion of woodlice with antenna lengths that are at least 0.23 inches, we subtract the proportion we found from 1 (since the total proportion is always 1).

So, 1 - 0.5793 = 0.4207

Therefore, approximately 42.07% of woodlice have antenna lengths that are at least 0.23 inches. Rounded to four decimal places, this is 0.4207.

Suppose that the antenna lengths of woodlice are approximately normally distributed with a mean of 0.22 inches and a standard deviation of 0.05 inches. What proportion of woodlice have antenna lengths that are at least 0.23 inches? Round your answer to at least four decimal places.

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