Question

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 264 days and a standard deviation of 10 days. a) What is the minimum pregnancy length that can be in the top 9% of pregnancy lengths? b) What is the maximum pregnancy length that can be in the bottom 7% of pregnancy lengths?

Answer 1

Answer:

Approximately 278 is the minimum pregnancy length that can be in the top 9% of pregnancy lengths.

approximately 279 is the minimum pregnancy length that can be in the top 7% of pregnancy lengths.

Step-by-step explanation:

We are given the following information:

Mean, μ = 264 days

Standard Deviation, σ = 10 days

We are given that the distribution of  pregnancy length in days is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(X>x) = 9% = 0.09

We have to find the value of x such that the probability is 0.09.

P(X > x)  

$P( X > x) = P( z > \displaystyle\frac{x - 264}{10})=0.09$  

$= 1 -P( z \leq \displaystyle\frac{x - 264}{10})=0.09$  

$P( z \leq \displaystyle\frac{x - 264}{10})=1-0.09=0.91$  

Calculation the value from standard normal z table, we have,  

$P(z \leq 1.341) = 0.91$

$\displaystyle\frac{x - 264}{10} = 1.341\\x = 277.41$  

Hence, approximately 278 is the minimum pregnancy length that can be in the top 9% of pregnancy lengths.

b)  P(X>x) = 7% = 0.07

We have to find the value of x such that the probability is 0.07.

P(X > x)  

$P( X > x) = P( z > \displaystyle\frac{x - 264}{10})=0.07$  

$= 1 -P( z \leq \displaystyle\frac{x - 264}{10})=0.07$  

$P( z \leq \displaystyle\frac{x - 264}{10})=1-0.07=0.93$  

Calculation the value from standard normal z table, we have,  

$P(z \leq 1.476) = 0.93$

$\displaystyle\frac{x - 264}{10} = 1.476\\x = 278.76$  

Hence, approximately 279 is the minimum pregnancy length that can be in the top 7% of pregnancy lengths.