C] The lengths of human pregnancies can be approximated by a normal distribution

Question

C] The lengths of human pregnancies can be approximated by a normal distribution with mean 280 days and standard deviation 17 days. A) Find the probability that a random sample of 36 pregnancies has mean gestation period between 272.5 and 273.5 days (the sample mean is one week before the mean due date); B) If a random sample of 100 pregnancies is selected, find the probability that the mean gestation period is no less than 275 days and does not exceed 285 days; C) If a random sample of 25 pregnancies is selected, find the probability that the mean gestation period is at most 270 days D) Find the probability that a random sample of 20 pregnancies has mean gestation period between 271.35 and 288.65 days. E) Repeat parts A), B), C), D) for one pregnancy selected at random. ANSWERS: A) 0.0070; B) 0.9968; C) 0.0016: D) 0.9774; E) 0.0220; 0.2282; 0.2776: 0.39 tudanto ore normal

Explanation / Answer

We are given

Mean = µ = 280

SD = = 17

Part A

We have to find P(272.5<Xbar<273.5)

We are given sample size n = 36

P(272.5<Xbar<273.5) = P(Xbar<273.5) – P(Xbar<272.5)

Z score for P(Xbar<273.5)

Z = (Xbar - µ) / [/sqrt(n)]

Z = (273.5 – 280) / [ 17/sqrt(36)]

Z =(273.5 – 280) / [ 17/6]

Z = -2.29412

P(Z<-2.29412) = 0.010892

P(Xbar<273.5) = 0.010892

Z score for P(Xbar<272.5)

Z = (272.5 – 280) / [ 17/sqrt(36)]

Z =(272.5 – 280) / [ 17/6]

Z = -2.64706

P(Z< -2.64706) = 0.00406

P(Xbar<272.5) = 0.00406

P(272.5<Xbar<273.5) = P(Xbar<273.5) – P(Xbar<272.5)

P(272.5<Xbar<273.5) = 0.010892 - 0.00406 = 0.006832

Required probability = 0.0068

Part B

Here, we have to find P(275<Xbar<285)

We are given n = 100

P(275<Xbar<285) = P(Xbar<285) – P(Xbar<275)

Z = (285 – 280)/ [ 17/sqrt(100) = 2.941176

P(Z< 2.941176) = 0.998365

Z = (275 – 280) / [17/sqrt(100)] = -2.94118

P(Z< -2.94118) = 0.001635

P(275<Xbar<285) = P(Xbar<285) – P(Xbar<275)

P(275<Xbar<285) = 0.998365 - 0.001635 = 0.99673

Required probability = 0.9967

Part C

We are given n = 25

We have to find P(Xbar<270)

Z = (270 – 280) / [17/sqrt(25)]

Z = -2.63158

P(Z< -2.63158) = 0.004249

Required probability = 0.001635

Part D

We are given n = 20

We have to find P(271.35<Xbar<288.65)

Z = (271.35 – 280) / [17/sqrt(20)] = -2.275528

P(Z< -2.275528) = 0.011437137

Z = (288.65 – 280) / [17/sqrt(20)]

Z = 2.275528

P(Z< 2.275528) = 0.988563

P(271.35<Xbar<288.65) = P(Xbar<288.65) – P(Xbar<271.35)

P(271.35<Xbar<288.65) = 0.988563 – 0.011437137 = 0.977126

Required probability = 0.977126

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