Question

17. Find the probability that a piece of data from a standard normal distribution will have a standard score described by the following:

a) Less than z = 1.25

b) Between z = – 2.03 and z = – 0.69

c) Find the score associated with the 62nd percentile

Answer 1

Solution

${\color{Blue} \mathbf{a)}}\;P(Less\;than\;z=1.25)\Rightarrow P(z<1.25)=?$

Refer Standard normal table/Z-table to find the probability or use excel formula "=NORM.S.DIST(1.25, TRUE)" to find the probability.

$P(z<1.25)=\mathbf{{\color{Red} 0.8944}}$

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${\color{Blue} \mathbf{b)}}\;P(Less\;than\;z=-2.03\;and\;z=-0.69)\Rightarrow P(-2.03<z<-0.69)=?$

$\Rightarrow P(z<-0.69)-P(z<-2.03)$

Refer Standard normal table/Z-table to find the probability or use excel formula "=NORM.S.DIST(-0.69, TRUE)" & "=NORM.S.DIST(-2.03, TRUE)" to find the probability.

$\Rightarrow 0.2451-0.0212$

$\Rightarrow \mathbf{{\color{Red} 0.2239}}$

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$\\{\color{Blue} \mathbf{c)}}\;62^{nd}\;percentile\;is\; an \;area\; of \;0.62 \;to\; the\; left\; of\; z\Rightarrow P(Z<z)=0.62$

Refer Standard normal table/Z-table, Lookup for z-score corresponding to area 0.62 to the left of the normal curve or use excel formula "=NORM.S.INV(0.62)" to find the z-score.

$P(Z<\mathbf{0.31})=0.62$

$\mathbf{score={\color{Red} 0.31}}$