I DO NOT have a graphing calculator or anything similar to a TI-84. I have a TI-30XIIS. If you can please explain another way to find the following problems without the TI-84, that includes using a Normal Distribution Table, that would be great.
P(0 ≤ Z ≤ 1.01)
P(x ≤ 40), μ = 50, σ = 10
(for this when I tried $$\frac{40-50}{10}$$ and it came out to -1, it was incorrect)
Find the area under the standard normal curve on the interval [-1.8, 1.8].
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$\begingroup$It is difficult to answer specifically because normal tables come in so many different varieties.
$P(0 \le Z \le 1.01).$ If you look in the margins of your normal table you may find a row for 1.0 and a column for .01. The corresponding value in the body of the table may be .3438, which is the probability you want. If it says .8438, then you have to subtract 0.5000, which is the probability below 0. There should be a picture at the head of the table with shading to tell you what kind of areas are in the table.
$P(X \le 40)$ for $X \sim \mathsf{Norm}(\mu = 50,\, \sigma = 10).$ $$P(X \le 40) - P\left(\frac{X-\mu}{\sigma} \le \frac{40-50}{10}\right) = P(Z \le -1) = 0.1587.$$ There is nothing wrong with getting a negative z-score. But you might not find negative z-scores in your table. Then you will have to use the symmetry of the standard normal distribution about 0 to get the answer. For example, if you find $P(0 \le Z \le 1) = 0.3413,$ you can use that to get .5 - 0.3413 = 0.1587.$
Consider the sketch of the standard normal density function below.
There are four areas separated by vertical red lines. Their areas represent probabilities 0.1587, 0.3413, 0.3413, and 0.1587, respectively. Added together these areas sum to 1. So you should be able to find the probability that $Z$ is in any interval with boundaries involving the z-scores -1, 0, and 1.
- $P(-1.8 \le Z \le 1.8) = 0.9282.$ In the sketch below the respective areas are 0.0359, 0.4641, 0.4641, and 0.0359. I will leave it to you to figure out how to get these areas from your particular printed tables, and how to put them together to get the answer.
Define a standard normal random variable $Z \sim \mathcal{N}(0,1)$. The area under the curve on the interval $[-\infty,x]$ will be $P(Z \le x)$.
Knowing this, try to convince yourself that
$$P(a \le Z \le b) = P(Z\le b) - P(Z\le a)$$
which will enable you to solve the first and last question with a normal distribution table. Also, in the first question, what do you know about $P(Z\le0)$?
For the second question, define another random variable $X\sim \mathcal{N}(50,10^{2})$. This random variable can be defined in terms of $Z$ in the following way:
$$X=50+10Z$$
Thus, $P(X \le 40)=P(50+10Z\le 40)$. Try to get this in the form $P(Z\le k)$, so you can use your normal distribution table.
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