Question

The lengths of a particular​ animal's pregnancies are approximately normally​ distributed, with mean mu equals280 days and standard deviation sigma equals12 days. ​(a) What proportion of pregnancies lasts more than 295 ​days? ​(b) What proportion of pregnancies lasts between 259 and 283 ​days? ​(c) What is the probability that a randomly selected pregnancy lasts no more than 262 ​days? ​(d) A​ "very preterm" baby is one whose gestation period is less than 250 days. Are very preterm babies​ unusual?

Answer 1

a)

Here, μ = 280, σ = 12 and x = 295. We need to compute P(X >= 295). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (295 - 280)/12 = 1.25

Therefore,
P(X >= 295) = P(z <= (295 - 280)/12)
= P(z >= 1.25)
= 1 - 0.8944 = 0.1056

b)

Here, μ = 280, σ = 12, x1 = 259 and x2 = 283. We need to compute P(259<= X <= 283). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (259 - 280)/12 = -1.75
z2 = (283 - 280)/12 = 0.25

Therefore, we get
P(259 <= X <= 283) = P((283 - 280)/12) <= z <= (283 - 280)/12)
= P(-1.75 <= z <= 0.25) = P(z <= 0.25) - P(z <= -1.75)
= 0.5987 - 0.0401
= 0.5586

c)

Here, μ = 280, σ = 12 and x = 262. We need to compute P(X <= 262). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (262 - 280)/12 = -1.5

Therefore,
P(X <= 262) = P(z <= (262 - 280)/12)
= P(z <= -1.5)
= 0.0668

d)

Here, μ = 280, σ = 12 and x = 250. We need to compute P(X <= 250). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (250 - 280)/12 = -2.5

Therefore,
P(X <= 250) = P(z <= (250 - 280)/12)
= P(z <= -2.5)
= 0.0062

yes it is unusual because probability is less than 0.05