Chapter 2.2, Problem 59E

(a)

To determine

Proportion of observations in a standard Normal distribution that satisfies the given statement.

Answer to Problem 59E

Around 0.9515 of the observations in a standard Normal distribution satisfy the statement.

Explanation of Solution

Given Information:

  $z>-1.66$

Calculations:

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -1.6 and the column that starts with .06 of the standard normal probability table for $P\left(z<-1.66\right)$ .

Then

Use the complement rule,

  $P\left(\text{not}A\right)=1-P\left(A\right)$

For probability

  $P\left(z>-1.66\right)$ .

  $\begin{array}{l}P\left(z>-1.66\right)=1-P\left(z<-1.66\right)\\ =1-0.0485\\ =0.9515\end{array}$

Therefore,

Around 0.9515 of the observations in a standard Normal distribution satisfy $z>-1.66$ .

(b)

To determine

Proportion of observations in a standard Normal distribution that satisfies the statement provided.

Answer to Problem 59E

Around 0.9493 of the observations in a standard Normal distribution satisfy the statement.

Explanation of Solution

Given Information:

  $-1.66<z<2.85$

Calculations:

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -1.6 and the column that starts with .06 of the standard normal probability table for $P\left(z<-1.66\right)$ .

And

See the row that starts with 2.8 and the column that starts with .05 of the standard normal probability table for $P\left(z<2.85\right)$ .

Then

The difference between the probabilities to the left of the boundaries is the probability between two boundaries.

  $\begin{array}{l}P\left(-1.66<z<2.85\right)=P\left(z<2.85\right)-P\left(z<-1.66\right)\\ =0.9978-0.0485\\ =0.9493\end{array}$

Therefore,

Around 0.9493 of the observations in a standard Normal distribution satisfy $-1.66<z<2.85$ .