Answer:
The answer is below
Step-by-step explanation:
Given that mean (μ) = 266 days, standard deviation (σ) = 16 days, sample size (n) = 16 women.
a) The mean of the sampling distribution of Xbar (
) is given as:
![\mu_x=\mu=266\ days](https://tex.z-dn.net/?f=%5Cmu_x%3D%5Cmu%3D266%5C%20days)
The standard deviation of the sampling distribution of Xbar (
) is given as:
![\sigma_x=\frac{\sigma}{\sqrt{n} } =\frac{16}{\sqrt{16} }=4](https://tex.z-dn.net/?f=%5Csigma_x%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%3D%5Cfrac%7B16%7D%7B%5Csqrt%7B16%7D%20%7D%3D4)
b) The z score is a measure in statistics used to determine by how much the raw score is above or below the mean. It is given by:
![z=\frac{x-\mu}{\sigma/\sqrt{n} }](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%20%7D)
For x > 270 days:
![z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{270-266}{\frac{16}{\sqrt{4} } }=1](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%20%7D%3D%5Cfrac%7B270-266%7D%7B%5Cfrac%7B16%7D%7B%5Csqrt%7B4%7D%20%7D%20%7D%3D1)
The probability the average pregnancy length exceed 270 days = P(x > 270) = P(z > 1) = 1 - P(z < 1) = 1 - 0.8413 = 0.1587 = 15.87%
Answer:
4
Step-by-step explanation:
Answer:
Option C - 2(-5)+2(3) ; -4
Step-by-step explanation:
Distributive property means breaking up the factors and multiplying them separately
So here we have 2(-5+3) = 2(-5)+2(3) = (-10)+(6) = (-4)
So (-4) is the value.
Hope this helps you.
492.9
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