1
$\begingroup$

I'm having trouble understanding the question. The solution looks simple, but I do not understand the logic behind the problem. For example, (a). How is it that z=1 when z=(x - mu)/(sigma). See problem image

Normal table

  • 1
    A few things: please change your question title to something more meaningful, and do not just post image links to your question. It will also help to draw a picture. – 2017-01-29
  • 0
    No wonder you're puzzled. The question has flaws. Will try to help, please see my Answer. – 2017-01-29

1 Answers 1

0

Let $X$ have a normal distribution with mean $\mu$ and standard deviation $\sigma.$ It is not possible to make printed distribution tables for all possible choices of $\mu$ and $\sigma.$ So we transform $X$ to $Z$ with the 'standardization' formula $Z = (X - \mu)/\sigma.$ If $X \sim \mathcal{Norm}(\mu = 100,\, \sigma=15)$ and you want $P(X \le 115),$ you write $$P(X \le 115) = P\left(\frac{X - \mu}{\sigma} \le \frac{115-100}{15} = 1\right) = P(Z \le 1) = 0.8413.$$ You can get $P(Z \le 1) = 0.8413$ from the printed standard normal table. In your table, find the entry for 1.0 in the left margin, and read 0.3413--to which you need to add 0.5000.

In the sketch below (following @SeanRobertson's suggestion), you want the area under the curve to the left of the dotted red line. The total area under the curve is $1,$ half on either side of the vertical green line. (Your table gives the area between the red and green lines.)

enter image description here

The answers to parts (a) and (b) are about 0.68 and about 0.95, respectively. Parts (c), (d) and (e) are (let me see how to put this delicately) total garbage as written. That's one of the reasons I'm trying to give you some help. (I guess that in (c) "$\mu$ to $\sigma$" was intended to mean "between $\mu$ and $\mu + \sigma."$)