Answer :
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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