The Normal Distribution

Question

z-Scores and Area under the Curve
A researcher collected data on adults using an intelligence test. The data for these adults were approximately normally distributed with the following mean and standard deviation		mean = 101101 points	standard deviation = 1717 points		Use this information and the online normal distribution calculator to answer the questions below. Round your percents to two decimal places.		Note: Use z-scores rounded to 2 decimal places and the standard normal distribution to avoid roundoff errors.
Normal Calculator
If an adult is randomly selected from the study, what is the probability the adult scored between 111111 points and 125125 points? %
If an adult is randomly selected from the study, what is the probability the adult scored less than 111.2111.2 points? %
If an adult is randomly selected from the study, what is the probability the adult scored more than 131.6131.6 points? %
If an adult is randomly selected from the study, what is the probability the adult scored less than 64.564.5 points or more than 142.3142.3 points? %

Answer 1

Hi Tra'von,

This question uses the classic equation in elementary statistics:

z=x-mu/sigma

x=value your question asks for

mu=mean

sigma=standard deviation

Note: You might want to be more careful with your decimals in the future. I know that intelligence tests do not typically have scores in the range of 101101, but that's what the number conveys. I suspect the actual mean is 101.101 and actual standard deviation 17.17, so that's what I'll go with.

mu=101.101

sigma=17.17

First Question

You want the probability that an adult scored between 111.11 and 125.12 points, so we need 2 z-scores. I will denote as z₁ and z₂ to avoid confusion:

Using equation from above:

x₁=111.11

mu=101.101

sigma=17.17

z₁=(111.11-101.101)/17.17

z₁=0.59

Now let's get z₂:

x₂=125.12

mu=101.101

sigma=17.17

z₂=(125.12-101.101)/17.17

z₂=1.40

Now, instead of P(111.11<x< z < 1.40) Recall that probabilities on the z-table are probabilities that z is less than the z-score, so first:

P(Z<0.59)=0.7224

P(Z<1.40)=0.9192

To compute the area between, simply do the subtraction:

P(0.59 < z < 1.40)= 0.9192 - 0.7224

P=0.1968

Second Question

This time, we only have one x:

x=111.2

mu=101.101

sigma=17.17

z=(111.2-101.01)/17.17

z=0.59

From z-table, P(z<0.59)=0.7224

P=0.7224

Third Question

This is similar to the second question, but we have an additional step at the end:

x=131.6

mu=101.101

sigma=17.17

z=(131.6-111.11)/17.17

z=1.19

P(Z<1.19)=0.8830

Now, the question asked for greater than, which means we must apply the Complement Rule, or what I call the "one minus trick:"

P(Z>a)=1-P(Z<a)

P(Z>1.19)=1-P(Z<1.19)

P(Z>1.19)=1-0.8830

P=0.1170

Final Question

Here, we need to treat the two cases separately and ultimately add the probabilities together. Again, I will use z₁ and z_2.

x₁=64.5645

mu=101.101

sigma=17.17

z₁=(64.5645-101.101)/17.17

z₁= -2.13

First part of the question asked for less than, so we can go to z-table and check:

P(z< -2.13)= 0.0166

Keep that value in mind; let's get z₂:

x₂=142.3

mu=101.101

sigma=17.17

z₂=(142.3-101.101)/17.17

z₂= 2.40

Now, we will go back to the z-table, but remember this question asked greater than, so:

P(Z<2.40)=0.9918

Remember the 1 minus trick:

P(Z>2.40)=1-P(Z<2.40)=

P(Z>2.40)=1-0.9918=0.0082

S, we add this to our probability above

P=0.0082 + 0.0166

P= 0.0248

I hope this helps.