11. Find z if the standard normal-curve area a) between 0 and z is 0.4726; b) to the left of z is 0.9868; c) to the left of z is 0.3085; d) between -z and z is 0.9282 12. If a random variable has the normal distribution with \mu = 82:0 and \sigma = 4:8, nd the probabilities that it will take on a value a) less than 89.2; b) greater than 78.4; c) between 83.2 and 88.0; d) between 73.6 and 90.4. 13. If the time to assemble an \easy to assemble" computer desk from a kit is a random variable having the normal distribution with \mu = 55:8 minutes and \sigma = 12:2 minutes, what are the probabilities that this kind of desk can be assembled in a) less than 49.7 minutes; b) anywhere from 61.9 and 74.1 minutes; c) more than 86.3 minutes? 14. With reference to Exercise 13, for what length of time is the probability 0.90 that one can assemble the desk in that many minutes or less?

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