Question

find the z value that corresponds to the given area.
in case you cant read the numbers on the graph. below are the numbers
a. 0.4013
b. 0.6700
c. 0.5910
d. 0.1977
problem 3

Explanation:

Step1: Use the standard normal distribution table (z - table)

The standard normal distribution table gives the cumulative probability \(P(Z\leq z)\) for different values of \(z\).

• For part a:

We know that \(P(Z\leq z)=0.4013 + 0.5\) (since the area to the left of \(z = 0\) is \(0.5\) in a standard normal distribution). But wait, no, actually, if the area is \(0.4013\) to the left of \(z\) (assuming it's on the left - hand side of the mean). Looking at the standard normal table, for \(P(Z\leq z)=0.4013\), we search for the value closest to \(0.4013\) in the body of the table. The \(z\) - value corresponding to \(P(Z\leq z)=0.4013\) is \(z=- 0.25\) (because in the standard normal table, for \(z=-0.25\), \(P(Z\leq - 0.25)=0.4013\)).

• For part b:

We look for the value closest to \(0.6700\) in the body of the standard normal table. The \(z\) - value corresponding to \(P(Z\leq z)=0.6700\) is \(z = 0.44\) (since for \(z = 0.44\), \(P(Z\leq0.44)=0.6700\)).

• For part c:

We look for the value closest to \(0.5910\) in the body of the standard normal table. The \(z\) - value corresponding to \(P(Z\leq z)=0.5910\) is \(z=0.23\) (since for \(z = 0.23\), \(P(Z\leq0.23)=0.5910\)).

• For part d:

Since the area is \(0.1977\) to the right of \(z\) (assuming it's on the right - hand side of the mean). We know that \(P(Z\leq z)=0.5 - 0.1977=0.3023\) (using the symmetry of the standard normal distribution \(P(Z\leq - z)=1 - P(Z\leq z)\)). Looking for \(0.3023\) in the standard normal table, we get \(z=-0.52\), and by symmetry, the \(z\) - value for the right - hand side (where the area to the right of \(z\) is \(0.1977\)) is \(z = 0.52\)

Answer:

a. \(z=-0.25\)
b. \(z = 0.44\)
c. \(z=0.23\)
d. \(z = 0.52\)