Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days and standard

Question

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days and standard deviation of 16 day

a. What is the probability that the sample mean length of pregnancy for 15 randomly selected women lasts less than 260 days.
b. The probability that the total length of pregnancy for 15 randomly selected women is greater than a certain amount is 0.05. Find this total length.
c. What is the probability that the sample total length of pregnancy for 7 randomly selected women is between 1800 and 1900 days?
d. The probability that the average length of pregnancy for randomly selected women is greater than 270 is 0.1151. Find the sample size of the random sample.

in progress 0
Abigail 2 weeks 2021-11-19T16:32:11+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-11-19T16:33:32+00:00

    Answer:

    Step-by-step explanation:

    Hello!

    X: length of human pregnancies from conception to birth.

    X~N(μ;σ²)

    μ= 264 day

    σ= 16 day

    If the variable of interest has a normal distribution, it’s the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

    X[bar] ~N(μ;σ²/n)

    When calculating a probability of a value of “X” happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

    When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

    a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

    P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

    b. P(X[bar]>b)= 0.05

    You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

    P(X[bar]≤b)= 0.95

    P(Z≤(b-μ)/(σ/√n))= 0.95

    The value of Z that accumulates 0.95 of probability is Z= 1.648

    Now we reverse the standardization to reach the value of pregnancy length:

    1.648= (b-264)/(16/√15)

    1.648*(16/√15)= b-264

    b= [1.648*(16/√15)]+264

    b= 270.81 days

    c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

    Symbolically:

    P(1800≤X[bar]≤1900) = P(X[bar]≤1900) – P(X[bar]≤1800)

    P(Z≤(1900-264)/(16/√7)) – P(Z≤(1800-264)/(16/√7))

    P(Z≤270.53) – P(Z≤253.99)= 1 – 1 = 0

    d. P(X[bar]>270)= 0.1151

    P(Z>(270-264)/(16/√n))= 0.1151

    P(Z≤(270-264)/(16/√n))= 1 – 0.1151

    P(Z≤6/(16/√n))= 0.8849

    With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

    P(Z≤1.200)= 0.8849

    Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }

    Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)

    (Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}

    Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}

    Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}

    n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2

    n = (16*(\frac{1.2}{(270-264)}))^2

    n= 10.24 ≅ 11 pregnant women.

    I hope it helps!

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )