Math  /  Data & Statistics

QuestionThe pregnancy durations (in days) for a population of new mothers can be approximated by a normal distribution, with a mean of 272 days and a standard deviation of 9 days. (a) What is the minimum pregnancy durations that can be in the top 8%8 \% of pregnancy durations? (b) What pregnancy durations would be considered unusual?

Studdy Solution

STEP 1

1. Pregnancy durations are normally distributed.
2. Mean (μ\mu) = 272 days.
3. Standard deviation (σ\sigma) = 9 days.
4. For part (a), we need to find the duration corresponding to the 92nd percentile.
5. For part (b), durations are considered unusual if they are more than 2 standard deviations away from the mean.

STEP 2

1. Calculate the minimum pregnancy duration for the top 8%.
2. Determine the range of durations considered unusual.

STEP 3

To find the minimum pregnancy duration in the top 8%, we need to find the 92nd percentile of the normal distribution.

STEP 4

Use the standard normal distribution table (or a calculator) to find the z-score that corresponds to the 92nd percentile. The z-score for the 92nd percentile is approximately 1.41.

STEP 5

Use the z-score formula to find the corresponding pregnancy duration: X=μ+z×σ X = \mu + z \times \sigma X=272+1.41×9 X = 272 + 1.41 \times 9

STEP 6

Calculate the value: X=272+12.69=284.69 X = 272 + 12.69 = 284.69

STEP 7

To find unusual pregnancy durations, calculate the range that is more than 2 standard deviations away from the mean: Lower bound=μ2σ=2722×9=254 \text{Lower bound} = \mu - 2\sigma = 272 - 2 \times 9 = 254 Upper bound=μ+2σ=272+2×9=290 \text{Upper bound} = \mu + 2\sigma = 272 + 2 \times 9 = 290
The minimum pregnancy duration in the top 8% is approximately 284.69 284.69 days.
Pregnancy durations considered unusual are those less than 254 254 days or greater than 290 290 days.

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