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:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
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)
[tex]P( X > x) = P( z > \displaystyle\frac{x - 264}{10})=0.09[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 264}{10})=0.09 [/tex]
[tex]P( z \leq \displaystyle\frac{x - 264}{10})=1-0.09=0.91 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z \leq 1.341) = 0.91[/tex]
[tex]\displaystyle\frac{x - 264}{10} = 1.341\\x = 277.41[/tex]
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)
[tex]P( X > x) = P( z > \displaystyle\frac{x - 264}{10})=0.07[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 264}{10})=0.07 [/tex]
[tex]P( z \leq \displaystyle\frac{x - 264}{10})=1-0.07=0.93 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z \leq 1.476) = 0.93[/tex]
[tex]\displaystyle\frac{x - 264}{10} = 1.476\\x = 278.76[/tex]
Hence, approximately 279 is the minimum pregnancy length that can be in the top 7% of pregnancy lengths.