Question

he pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of

267267

days and a standard deviation of

88

days.​(a) What is the minimum pregnancy length that can be in the top

99​%

of pregnancy​ lengths?​(b) What is the maximum pregnancy length that can be in the bottom

33​%

of pregnancy​ lengths?

​(a) The minimum pregnancy length is

nothing

days.

​(Round to one decimal place as​ needed.)

show steps plz

Score: 0 of 1 pt 10 of 12 (8 complete) ► HW Score: 46.15%, 6 of 1... 5.3.38-T Assigned Media Question Help The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 267 days and a standard deviation of 8 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 3% of pregnancy lengths? (a) The minimum pregnancy length is days. (Round to one decimal hace as needed.) Enter your answer in the answer box and then click Check Answer

Answer 1

Let X denotes the pregnancy length for a randomly selected mother.

X ~ Normal(267, 8²)

a) To find x₁ such that

$\Rightarrow 1 - P(X\leq x_1) = 0.09$

$\Rightarrow P(X\leq x_1) =1 - 0.09 = 0.91$

$\Rightarrow P(\frac{X-267}{8}\leq \frac{x_1-267}{8}) = 0.91$

$\Rightarrow P(Z \leq \frac{x_1-267}{8}) = 0.91$

$\Rightarrow \Phi(\frac{x_1-267}{8}) = 0.91$

$\Rightarrow \frac{x_1-267}{8} = \Phi^{-1}(0.91) = 1.34076$

$\Rightarrow x_1= 267 +(1.34076*8)$

$\Rightarrow x_1= 277.726$

$\Rightarrow x_1 \approx 277.7$

b)

To find x₂ such that

$P(X\leq x_2) = 0.03$

$\Rightarrow P(\frac{X-267}{8}\leq \frac{x_2-267}{8}) = 0.03$

$\Rightarrow P(Z \leq \frac{x_2-267}{8}) = 0.03$

$\Rightarrow \Phi(\frac{x_2-267}{8}) = 0.03$

$\Rightarrow \frac{x_2-267}{8} = \Phi^{-1}(0.03) = -1.88079$

$\Rightarrow x_2 = 267-(1.88079*8)$

$\Rightarrow x_2 = 251.95368$

$\Rightarrow x_2 \approx 252.0$