Mean and standard deviation and probability

[tcardwe3]

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

 

[MarkFL]

[box=red]

Sampling Distribution of a Normal Variable​

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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?

 

[pickslides]

For the second part of the question.

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