Find the value of Z under the Standard Normal Curve

Question

I dont understand this question:

1. Between $Z$ and $1.52$ is $0.1306$

Question:

If $Z$ is has a value of $0$ and what is the value for $1.52$? Is it the value that corresponds to the Areas under the normal curve which is $0.9357$? Or is the value for $1.52$ is $0.1306$?

Answer 1

I think you want to find Z such that the area under the standard normal curve between the points Z and 1.52 is 0.1306 or 13.06%. For example we know that the area under the standard normal curve between z=0 and infinity is exactly 0.5 or 50%

Z should be 0.86. You can check the are under the standard normal curve from Z=0.86 to 1.52 is 0.1306:

http://onlinestatbook.com/2/calculators/normal_dist.html

Answer 2

Formally it is asked for the value $z_u$ that fullfills the following equation:

$P(z_u\leq Z\leq 1.52)=0.1306$

In words: The probability that the random variable $Z$ is between $z_u$ and $1.52$ is $0.1306$, where $Z$ is standard normally distributed What is the value of $z_u$?

$P(z_u\leq Z\leq 1.52)=P(Z\leq 1.52)-P(Z \leq z_u)=0.1306$

Note $z_u$ is a constant and not random variable. It can be denoted by other characters as well.

$P(Z\leq 1.52)$ is the area under the normal curve from $-\infty$ to $1.52$

$P(Z\leq z_u)$ is the area under the normal curve from $-\infty$ to $z_u$

We have to calculate the sum of the red and blue area (1+2). This is $P(Z\leq 1.52)=\Phi(1.52)$, where $\Phi(z)$ is the (cummulative) function of the standard normal distribution. And then substract the red area (1): $P(Z\leq z_u)=\Phi(z_u)$

$\Phi(1.52)-\Phi(z_u)=0.1306$

You´re right that $\Phi(1.52)=0.936$. Therefore the equation becomes

$0.936-\Phi(z_u)=0.1306$

Solving the equation for $\Phi(z_u)$ gives $\Phi(z_u)=0.8054$

Using this calculator we get $z_u=0.861$