Chapter 11, Problem 11.8AYK

(a)

To determine

To use the technology of your choice to find the proportion of observations from a standard normal distribution and also sketch a standard normal curve and shade the area.

Answer to Problem 11.8AYK

The proportion of observations for $z<2.85$ is $0.99781$ .

Explanation of Solution

It is given that we have to use the technology of choice to find the proportion of observations from a standard normal distribution so, we will use excel function to solve the probability. The function used is as:

  $=\text{NORM}\text{.S}\text{.DIST(}z,\text{cumulative)}$

Thus, we have to find the probability for $z<2.85$ , then the calculation is as:

For z<2.85

=NORM.S.DIST(2.85,TRUE)

The result is as:

For z<2.85

0.99781

Therefore, the proportion of observations for $z<2.85$ is $0.99781$ .

And the sketch for the following is as:

  

(b)

To determine

To use the technology of your choice to find the proportion of observations from a standard normal distribution and also sketch a standard normal curve and shade the area.

Answer to Problem 11.8AYK

The proportion of observations for $z>2.85$ is $0.002186$ .

Explanation of Solution

It is given that we have to use the technology of choice to find the proportion of observations from a standard normal distribution so, we will use excel function to solve the probability. The function used is as:

  $=\text{NORM}\text{.S}\text{.DIST(}z,\text{cumulative)}$

Thus, we have to find the probability for $z>2.85$ , then the calculation is as:

For z>2.85

=1-NORM.S.DIST(2.85,TRUE)

The result is as:

For z>2.85

0.002186

Therefore, the proportion of observations for $z>2.85$ is $0.002186$ .

And the sketch for the following is as:

  

(c)

To determine

To use the technology of your choice to find the proportion of observations from a standard normal distribution and also sketch a standard normal curve and shade the area.

Answer to Problem 11.8AYK

The proportion of observations for $z>-1.66$ is $0.95154$ .

Explanation of Solution

It is given that we have to use the technology of choice to find the proportion of observations from a standard normal distribution so, we will use excel function to solve the probability. The function used is as:

  $=\text{NORM}\text{.S}\text{.DIST(}z,\text{cumulative)}$

Thus, we have to find the probability for $z>-1.66$ , then the calculation is as:

For z>-1.66

=1-NORM.S.DIST(-1.66,TRUE)

The result is as:

For z>-1.66

0.95154

Therefore, the proportion of observations for $z>-1.66$ is $0.95154$ .

And the sketch for the following is as:

  

(d)

To determine

To use the technology of your choice to find the proportion of observations from a standard normal distribution and also sketch a standard normal curve and shade the area.

Answer to Problem 11.8AYK

The proportion of observations for $-1.66<z<2.85$ is $0.94936$ .

Explanation of Solution

It is given that we have to use the technology of choice to find the proportion of observations from a standard normal distribution so, we will use excel function to solve the probability. The function used is as:

  $=\text{NORM}\text{.S}\text{.DIST(}z,\text{cumulative)}$

Thus, we have to find the probability for $-1.66<z<2.85$ , then the calculation is as:

For -1.66

=NORM.S.DIST(2.85,TRUE)-NORM.S.DIST(-1.66,TRUE)

The result is as:

For -1.66

0.949356

Therefore, the proportion of observations for $-1.66<z<2.85$ is $0.94936$ .

And the sketch for the following is as:

  

(e)

To determine

To explain in your words why these proportions can also be considered probabilities.

Explanation of Solution

As we know that these are the probabilities as they satisfy all the three conditions that are:

For any event A $P(A)\ge 0$ .

Probability of sample space i.e. total probability is one and total area under curve is also one.

If $A_1,A_2,A_3$ are disjoint events then $P(A_{1}U{A}_{2}U{A}_{3}U\mathrm{....})=P(A_1)+P(A_2)+P(A_3)+\mathrm{..}$ .

Thus, if all the above three conditions are satisfied then these proportions can also be considered probabilities.