Question

T2.8. Which of the following is not correct about a standard Normal distribution?
(a) The proportion of scores that satisfy $0<z<1.5$is $0.4332$.
(b) The proportion of scores that satisfy $z<-1.0$is $0.1587$.
(c) The proportion of scores that satisfy $z>2.0$is $0.0228$.
(d) The proportion of scores that satisfy $z<1.5$is $0.9332$.
(e) The proportion of scores that satisfy$z>-3.0$is $0.9938$.

Short answer

Option (e) is a conventional Normal distribution that is incorrect; the proportion of scores that fulfil $z>-3.0$ is$0.9938.$

Step-by-step solution

Given information

(a) The proportion of scores that satisfy $0<z<1.5$ is $0.4332$.
(b) The proportion of scores that satisfy $z<-1.0$ is $0.1587$.
(c) The proportion of scores that satisfy $z>2.0$ is $0.0228$.
(d) The proportion of scores that satisfy $z<1.5$ is $0.9332$.
(e) The proportion of scores that satisfy $z>-3.0$ is $0.9938$.

Explanation

(a) In the normal probability table in the appendix, the proportion of scores less than $0$is shown in the row with $0.0$and the column with $.00$.

$P(z<0)=0.5000$

In the normal probability table in the appendix, in the row with $1.5$and in the column with $.00$, the fraction of scores less than $1.5$is shown.

$P(z<1.5)=0.9332$

The difference between the proportions of scores to the left of two $z-$scores is the proportion of scores between the two $z-$scores.

localid="1649920681288" $$\begin{gathered} P(0<z<1.5)=P(z<1.5)-P(z<0) \\ =0.9332-0.5000 \\ =0.4332 \end{gathered}$$

Hence, the proportion of scores that satisfy $0<z<1.5$is $0.4332$So, the given statement is correct.

Command $Ti83/84$-calculator: Normalcdf $(0,1.5,0,1)$

Explanation

(b) The proportion of scores lower than $-1.0$is given in the normal probability table in the appendix in the row with $-1.0$and in the column with $.00$
$P(z<-1.0)=0.1587$
Hence, the proportion of scores that satisfy $z<-1.0$is $0.1587$. So, the given statement is correct.
Command $Ti83/84$-calculator: Normalcdf$(-1E99,-1.0,0,1)$

Explanation

(c) The proportion of scores lower than $2.0$is given in the normal probability table in the appendix in the row with $2.0$and in the column with $.00$
$P(z<2.0)=0.9772$
The total probability needs to be$1$, thus the probability to the right of $2.0$is $1$decreased by the probability to the left of $2.0$
$$\begin{gathered} P(z>2.0)=1-P(z<2.0) \\ =1-0.9772 \\ =0.0228 \end{gathered}$$

Hence, the proportion of scores that satisfy$z>2.0$is $0.0228$. So, the given statement is correct.

Command $Ti83/84$-calculator: Normalcdf $(2.0,1E99,0,1)$

Explanation

(d) The proportion of scores lower than $1.5$is given in the normal probability table in the appendix in the row with $1.5$and in the column with $.00$
$P(z<1.5)=0.9332$
Hence, the proportion of scores that satisfy $z<1.5$is $0.9332$. So, the given statement is correct.
Command $Ti83/84$-calculator: Normalcdf$(-1E99,1.5,0,1)$.

Explanation

(e) The proportion of scores lower than $-3.0$is given in the normal probability table in the appendix in the row with $-3.0$and in the column with$.00$
$P(z<-3.0)=0.0013$
The total probability needs to be $1$, hence the probability to the right of $2.0$is $1$decreased by the probability to the left of $2.0$
$$\begin{gathered} P(z>-3.0)=1-P(z<-3.0) \\ =1-0.0013 \\ =0.9987 \end{gathered}$$

Hence, the proportion of scores that satisfy $z>-3.0$is$0.9987$. So, the given statement is not correct (as $0.9987$is different from

$0.9938$).

Command $Ti83/84$-calculator: Normalcdf$(-3.0,1E99,0,1)$.

So, the option (e) is correct.