Question

Women’s weights are normally distributed with a mean given by µ = 143 lb and a standard deviation given by σ = 29 lb. Find the third decile, D3, which separates the bottom 30% from the top 70%._______

Answer 1

The third decile (D3) for women's weights, separating the bottom 30% from the top 70%, can be found by converting the 30th percentile z-score to an actual value using the mean and standard deviation, which results in approximately 127.92 pounds.

Step-by-step explanation:

To find the third decile (D3) for women's weights, which separates the bottom 30% from the top 70%, we need to use the mean (µ = 143 lb) and the standard deviation (σ = 29 lb) for the normally distributed weights. The third decile corresponds to the 30th percentile in the standard normal distribution.

First, we use a z-table to find the z-value that corresponds to the 30th percentile. Looking at standard normal distribution tables, or using a calculator, we find that the z-score corresponding to the 30th percentile is approximately -0.52.

Then, we use the formula for converting a z-score to an actual value in the distribution:

X = µ + z × σ

Substituting our values, we get:

X = 143 lb + (-0.52 × 29 lb)

X = 143 lb - 15.08 lb

X = 127.92 lb

Therefore, the third decile for women's weights is about 127.92 lb.

Answer 2

127.7924

Step-by-step explanation:

Given that women’s weights are normally distributed with a mean given by µ = 143 lb and a standard deviation given by σ = 29 lb.

Let x be the weight of women

X is N (143, 29)

Or
`(x-143)/(29)` is N(0,1)

From standard normal distribution table we can find D3 i.e. 30th percentile and then convert suitably to X score

30th percentile in Normal =-0.5244

Corresponding X = Mean -0.5244 * std deviation

=
`143-0.5244*29\\=143-15.2076\\=127.7924`

The third decile, D3, which separates the bottom 30% from the top 70%.___is 127.7924____