MHB Mean and standard deviation and probability

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The discussion revolves around calculating the mean and standard deviation for a sample of systolic blood pressure among women aged 18-24, which is normally distributed with a mean of 114.8 and a standard deviation of 13.1. The correct mean and standard deviation for the sample of 100 women is 114.8 and 1.31, respectively, as the standard deviation of the sample mean is calculated using the formula σ/√n. For the probability that the mean systolic blood pressure is between 112.2 and 116.4, participants are guided to use the Z-score formula for normal distributions. The discussion emphasizes understanding sampling distributions and the application of probability concepts. This foundational knowledge is crucial for solving similar statistical problems effectively.
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I did the problem but none of my answers match up with the answer choices so I'm obviously doing it wrong. Can someone show me how to do these two problems. I have a test coming up and I am so behind

For women aged 18-24, the systolic blood pressure (in hg mm) are normally distributed with a mean of 114.8 and a standard deviation of 13.1
100 women between 18 and 24 are randomly selected, let t represent the systolic pressure of 100 women
-Find the mean and standard deviation of t
a) 114.8, 131
b) 114.8, 13.1
c) 114.8, 1.31
d) 11.48, 1.31
d) 11.48, 1.31

-What is the probability that the mean systolic blood pressure t is between 112.2 and 116.4
a) .8649
b) .3879
c) .1578
d) .571
e) .0324
 
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Sampling Distribution of a Normal Variable

Given a random variable $$X$$. Suppose that the population distribution of $$X$$ is known to be normal, with mean $\mu$ and variance $\sigma^2$, that is, $X\sim N(\mu,\sigma)$. Then, for any sample size $n$, it follows that the sampling distribution of $X$ is normal, with mean $\mu$ and variance $\dfrac{\sigma^2}{n}$, that is, $\overline{X}\sim N\left(\mu,\dfrac{\sigma}{\sqrt{n}}\right)$[/box]

Based on this, what would you say the correct answer to the first part of the problem is?
 
For the second part of the question.

$\displaystyle P\left(112.2<\bar{T}<116.4\right) = P\left(\frac{112.2-\mu}{\frac{\sigma}{\sqrt{n}}}<Z<\frac{116.4-\mu}{\frac{\sigma}{\sqrt{n}}}\right)= \cdots$
 
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