Use the table to answer the question. Note: Round z-scores to the nearest hundredth and then find the required A values using the table.
A biologist found the wingspans of a group of monary butterflies to be normally distributed with a mean of 54.7 mm and a standard deviation of 2.4 mm. What percent of the butterflies had the following wingspans? (Round your answers to one decimal place.)
(a) less than 48.8 mm
%
(b) between 49 and 56 mm
To find the required A values using the z-score table, we first need to calculate the z-scores for the given wingspan values.
(a) To find the percentage of butterflies with wingspans less than 48.8 mm, we need to calculate the z-score for this value.
First, subtract the mean from the given value and divide it by the standard deviation:
z = (48.8 - 54.7) / 2.4
z ≈ -2.46
Next, we round the z-score to the nearest hundredth:
z ≈ -2.46
Now, we need to find the required A value for this z-score using the z-score table.
From the table, we see that the closest z-score value to -2.46 is -2.5, and the corresponding A value is 0.0062. Therefore, the percentage of butterflies with wingspans less than 48.8 mm is approximately 0.62%.
(b) To find the percentage of butterflies with wingspans between 49 and 56 mm, we need to find the corresponding z-scores for both values.
For 49 mm:
z = (49 - 54.7) / 2.4
z ≈ -2.38
For 56 mm:
z = (56 - 54.7) / 2.4
z ≈ 0.54
Next, we round the z-scores to the nearest hundredth:
For 49 mm: z ≈ -2.38
For 56 mm: z ≈ 0.54
Now, we need to find the required A values for these z-scores using the z-score table.
From the table, we find:
For z ≈ -2.38, the corresponding A value is 0.0078.
For z ≈ 0.54, the corresponding A value is 0.7054.
To find the percentage of butterflies with wingspans between 49 and 56 mm, we subtract the A value for 49 mm from the A value for 56 mm:
A(49 mm) - A(56 mm) = 0.7054 - 0.0078 = 0.6976
Therefore, the percentage of butterflies with wingspans between 49 and 56 mm is approximately 69.8%.
To find the required A values using the z-score table, we need to first calculate the z-scores for the given wingspans. The z-score formula is:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the given value
- μ is the mean
- σ is the standard deviation
(a) Let's calculate the z-score for 48.8 mm:
z = (48.8 - 54.7) / 2.4
z = -2.46
Now, we can find the A value corresponding to the z-score of -2.46 using the z-score table. The z-score table provides the area under the standard normal distribution curve.
From the table, we can see that the closest z-score to -2.46 is -2.5. The area corresponding to a z-score of -2.5 is 0.0062.
Therefore, the percentage of butterflies with wingspans less than 48.8 mm is approximately 0.62%.
(b) To find the percentage of butterflies with wingspans between 49 and 56 mm, we need to calculate the z-scores for both values.
For 49 mm:
z = (49 - 54.7) / 2.4
z = -2.38
For 56 mm:
z = (56 - 54.7) / 2.4
z = 0.54
Now, we can find the A values corresponding to these z-scores using the z-score table.
The closest z-score to -2.38 is -2.4, with an area of 0.0082.
The closest z-score to 0.54 is 0.5, with an area of 0.7040.
To find the percentage between these two values, we subtract the smaller area from the larger area:
0.7040 - 0.0082 = 0.6958
Therefore, the percentage of butterflies with wingspans between 49 and 56 mm is approximately 69.6%.