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

Answer:

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 $\frac{x-143}{29}$ 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____

Answer 2

The third decile (D3) will be "127.79".

Given values:

• Mean, $\mu = 143$
• Standard deviation, $\sigma = 29$

We know,

• $P(X<x) = 0.3$

Now,

→ $z = \frac{(x- \mu)}{\sigma}$

     $= \frac{(x-143)}{29}$

then,

→ $P[Z<(\frac{x-143}{29} )] =0.3$

→              $\frac{(x-143)}{29} =-52$

By using the normal table, we get

→                      $x = 127.79$ (third decile)

Thus the above response is right.  

Learn more about probability here:

https://brainly.com/question/19341630