Question

Lila, a marine biologist, calculated the mean and

standard deviation s for the weights of shrimp in a
sample of 1200 shrimp. She wanted to test how well the
weights fit the 68-95-99.7 rule for normal distributions.
Here are the results:
Bucket
x≤-28T
- 2sz ≤x≤ - Sz
-Sz≤x≤ã
≤x≤ + Sz
+ S₂ ≤ x ≤Ã+2s
x ≥ã+2sT
Expected
2.5%
Expected:
13.5%
34%
34%
13.5%
2.5%
# of shrimp
Lila wants to perform a x² goodness-of-fit test to
determine if these results suggest that the actual
distribution of shrimp doesn't match the expected
distribution.
40
140
400
450
110
60
What is the expected count of shrimp in the bucket
≤x≤+s, in Lila's sample?
You may round your answer to the nearest hundredth.

Answer 1

For a sample of 1200 shrimp, the expected count within one standard deviation from the mean (≤x≥+s) is 34% of 1200, which equals 408 shrimp.

Step-by-step explanation:

Lila, the marine biologist, has 1200 shrimp in her sample, and she expects the weights within one standard deviation of the mean to account for 34% of her sample.

This is part of applying the 68-95-99.7 rule, which states that for a normally distributed dataset, 68% of the data should fall within one standard deviation from the mean.

To calculate the expected count of shrimp in the bucket ≤x≥+s, we use the following steps:

1. Calculate 34% of the total sample size: 0.34 × 1200.
2. Apply the result from step 1 to find the expected count.

Therefore, the expected count of shrimp in the considered bucket is:

= 0.34 × 1200

= 408 shrimp.