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## Here's the Solution to this Question

11.

We shall obtain the values of z from the standard normal tables.

$a)$

z=1.92

$b)$

z=2.22

$c)$

z=-0.5

$d)$

If the area between -z and z is 0.9282 it implies that the area to the left of -z is ${1-0.9282\over2}=0.0359$. Thus $-z=-1.8\implies z=1.8$

12.

$\mu=82\\\sigma=4.8$

$a)$

$p(x\lt 89.2)=p(Z\lt{89.2-82\over4.8})=p(Z\lt 1.5)=\phi(1.5)=0.9332$

$b)$

$p(x\gt78.4)=p(Z\gt {78.4-82\over 4.8})=p(Z\gt -0.75)=1-p(Z\lt -0.75)=1-0.2266=0.7734$

$c)$

$p(83.2\lt x\lt 88)=p({83.2-82\over 4.8}\lt Z\lt{88-82\over 4.8})=p(0.25\lt Z\lt 1.25)=\phi(1.25)-\phi(0.25)=0.8944-0.5987=0.2957$

$d)$

$p(73.6\lt x\lt 90.4)=p({73.6-82\over4.8}\lt Z\lt{90.4-82\over4.8})=p(-1.75\lt Z\lt 1.75)=\phi(1.75)-\phi(-1.75)=0.9599-0.0401=0.9198$

13.

$\mu=55.8\\\sigma=12.2$

$a)$

$p(x\lt 49.7)=p(Z\lt {49.7-55.8\over 12.2})=p(Z\lt -0.5)=0.3085$

$b)$

$p(61.9\lt x\lt 74.1)=p({61.9-55.8\over12.2}\lt Z\lt{74.1-55.8\over12.2})=p(0.5\lt Z\lt 1.5)=\phi(1.5)-\phi(0.5)=0.9332-0.6915=0.2417$

$c)$

$p(x\gt 86.3)=p(Z\gt {86.3-55.8\over12.2})=p(Z\gt 2.5)=1-p(Z\lt 2.5)=1-0.9938=0.0062$

14.

Here, we find the value $y$ such that $p(x\lt y)=0.90$

Now,

$p(x\lt y)=p(Z\lt {y-55.8\over12.2})=0.90$. This implies that, $\phi({y-55.8\over 12.2})=0.90\implies {y-55.8\over 12.2}=Z_{0.90}$ where, $Z_{0.90}$ is the table value associated with 0.90. From tables, $Z_{0.90}=1.281552$

So,

${y-55.8\over 12.2}=1.281552\implies y=71.4349344$

Therefore, the probability that one can assemble the desk in less than 71.4349344 minutes is 0.90