Question

The duration (in days) of human pregnancies follows approximately the N(266,16) distribution. Reference: Ref 3-9 How many days would a human pregnancy need to last to be among the top 10% of all durations? Question 5Select one: a. 239.68 days b. 245.52 days c. 286.48 days d. 292.32 days

Answer 1

The duration of a pregnancy in the top 10% of all cases, using a normal distribution with mean 266 days and standard deviation 16 days, is approximately 286.48 days correct option is c.

Given parameters

Mean (µ) = 266 days

Standard deviation (σ) = 16 days

Find the z-score for the top 10% (90th percentile).

The z-score formula is:

`\[ z = \frac{{X - \mu}}{{\sigma}} \]`

For the 90th percentile (top 10%), we need to find the z-score corresponding to this percentile using a standard normal distribution table or calculator. The z-score for the 90th percentile is approximately 1.28.

Use the z-score formula to solve for the duration
`\(X\)`:

Given:

`\[ z = 1.28 \]`

`\[ \mu = 266 \]`

`\[ \sigma = 16 \]`

The formula is:

`\[ z = \frac{{X - \mu}}{{\sigma}} \]`

Substitute the values:

`\[ 1.28 = \frac{{X - 266}}{{16}} \]`

Solve for
`\(X\)` (the duration of pregnancy):

Multiply both sides by 16:

`\[ 1.28 * 16 = X - 266 \]`

`\[ 20.48 = X - 266 \]`

Add 266 to both sides:

`\[ X = 266 + 20.48 \]`

X ≈ 286.48

So, the duration of a human pregnancy that would place it among the top 10% of all durations is approximately
`\(286.48\)` days, which corresponds to option c.