Question

In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.) Ф(c)-0.9838 (a) (b) P(O SZ sc) 0.2967 (c) PC s Z)0.1230 (d) P(-c c)-0.6424 Z (e) P(c s IZ1)-0.0160

Answer 1

SOLUTION

(a) (1)(c) 0.9838

2.14

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(b) P(0 < Z < c) = 0.2967

P(Z < c)-P(Z < 0) 0.2967

P(Z < c) _ 0.5 0.2967

P(Z < c) = 0.7967

c = 0.83

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(c) P(с < Z) 0.1230

с 1.16

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(d) P(-c < Z < c) 0.6424

Assume the mean to be symmetrically distributed

0.1788 for two tail test, _-

from Z-table, Lookup fro Z-value corresponding to area 0.1788 to the left of the normal curve.

P(Z <ー0.92) 0.1788

from Z-table, Lookup fro Z-value corresponding to area 0.1788 to the right of the normal curve.

$\Rightarrow P(Z>\mathbf{0.92})=0.1788$​​​​​​​

$\therefore P(-0.92<Z<0.92)=0.6424$

$\therefore c=\mathbf{{\color{Red} 0.92}}$

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${\color{Blue} \mathbf{(e)}}\;P(c\leq \mid Z\mid )=0.0160$

$P(Z\leq c)= 1-\left \{ \frac{0.0160}{2} \right \}$

$P(Z\leq c)=\mathbf{0.9920}$

$\therefore c=\mathbf{{\color{Red} 2.41}}$