1

Evaluate the following for a standard normal probability distribution, drawing an appropriate picture.

A.) Find the standard normal z-value so that the area that lies to the right of z is 0.2119.

For part A, I did P(Z > 0.8) = 1 - P(Z < 0.8) = 0.2119 and shaded everything to the right of 0.8 on my drawing.

B.) Find the standard normal z-value so that the area between –z and z is 0.9030.

My professor ended lecture early and did not cover this material but I had this question for homework. For part A I am not sure if I am on the right track as far as my thought process and answer go. On part B I am completely lost. Any help would be much appreciated.

user102156
  • 445
  • 1
    A) is right. And the area between $-z$ and $z$ is $2\Phi(z)-1=0.9030$ – callculus42 Dec 06 '21 at 17:37
  • You are on the right track for A, and @callculus42 has given you the key clue for B. In my Answer I show how you can use R (or other software), but you can also use a printed standard normal CDF table. But the main point of by Answer is to show you the plots. // These kinds of computations are important as you use the normal distribution to solve problems. It is a good idea to start by making at least a rough sketch as a guide to use of normal tables. – BruceET Dec 08 '21 at 07:48

1 Answers1

1

As @Callculus42 has commented, you already have the answer to Part A. In R statistical software, where qnorm is the standard normal quantile function (invese CDF), we see that $P(Z > 0.8) = 0.2110.$ I hope you know how to get this result from a printed table of the standard normal CDF.

qnorm(1-.2119)
[1] 0.799846

This is the area under the standard normal density curve to the right of the solid vertical black line.

enter image description here

R code for figure:

hdr = "Standard Normal Density"
curve(dnorm(x), -3, 3, lwd=2, col="blue", 
      ylab="Density", xlab="z", main=hdr)
 abline(h=0, col="green2")
 abline(v=0, col="green2")
 abline(v=.8)

For Part B, the lower boundary $-1.659575$ should cut probability $(1 - .9030)/2 = 0.0485$ from the left tail of the standard normal distribution and (by symmetry) the upper boundary $1.659575$ should cut the same probability from the right tail.

qnorm(0.0485)
[1] -1.659575

Thus $P(-1.659575 \le Z \le 1.659575) = 0.9030.$

diff(pnorm(c(-1.659575,1.659575)))
[1] 0.903

In the figure below, $0.9030$ is the area under the standard normal curve between the dotted red lines at $\pm 1.659575.$

enter image description here

hdr = "Standard Normal Density"
curve(dnorm(x), -3, 3, lwd=2, col="blue", 
      ylab="Density", xlab="z", main=hdr)
 abline(h=0, col="green2")
 abline(v=0, col="green2")
 abline(v = c(-1.659575,1.659575), col="red", lwd=2, lty="dotted")

Note: Both of these computations are of types you will need to do repeatedly as you learn to use the standard normal distribution in probability and statistics. It is a good idea to begin by making sketches.

BruceET
  • 51,500