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The length of pregnancies are normally distributed with a mean of 208 days and a standard deviation of 15 days a. Find the probability of a pregnancy lasting 308 days or longer, b. ll the length of pregnancy is in the lowest 3%, then the baby is
premature. Find the length that separates premature babies from those who are not premature.


a. The probability that a pregnancy wit last 308 days or longer is II
(Round to four decimal places as needed)

Answer :

Fichohgo

The required probability values using the normal distribution concept are :

  • P(X ≥ 308) = 0

  • Length seperating premature babies = 179.785 days.

Given the Parameters :

  • Mean, μ = 208
  • Standard deviation, σ = 15

Recall : Zscore = [(X - μ) ÷ σ]

The probability of a pregnancy lasting 308 days or longer :

Zscore = (308 - 208) ÷ 15 = 6.667

P(Z ≥ 6.667) = 1 - P(Z ≤ 6.667)

Using a normal distribution table

P(Z ≤ 6.667) = 1

P(Z ≥ 6.667) = 1 - 1 = 0

The lowest 3% :

The Zscore corresponding to the lower 3% using the normal distribution table is :

P(Z < 0.03) ; Zscore = - 1.881

Calculate the value of X

Zscore = [(X - μ) ÷ σ]

-1.881 = (X - 208) / 15

-28.215 = X - 208

X = - 28.215 + 208

X = 179.785 days

Therefore, length of premature babies are less than or equal to 179.785 days (≤ 179.785 days)

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