Question

(1 point) Suppose z is a normally distributed random variable with u = 10.6 and o = 1.6. Find each of the following probabilities: (a) P(8.6 << < 15.1) (b) P(5.1 5*< 15.8) (c) P(7.1 Su 15.9) (d) P(x > 5.3) = (e) P(x< 13.4) =

Answer 1

Answer:-

Given that:-

u= 10.6, 0 = 1.6

Formula : $Z=\frac{X-\mu}{\sigma}$

(a)
X-1 8.6 – 10.6 P(8.6 << < 15.1) = PE- 1.6 < < 15.1 – 10.6 1.6 o
$=P[-1.25\leq Z\leq2.8125 ]$

  

= P(2.8125) - P(-1.25

== 0.9975-0.1057

  

  

b)$P(5.1\leq x\leq 15.8)=P[\frac{5.1-10.6}{1.6}\leq \frac{X-\mu}{\sigma }\leq \frac{15.8-10.6}{1.6}]$

P(-3.4375 < 3 <3.25)

  $=P(3.25)-P(-3.4375)$

  $=0.9994-0.0003$

$=0.9991$

c)$P(7.1\leq x\leq 15.9)=P[\frac{7.1-10.6}{1.6}\leq \frac{X-\mu}{\sigma }\leq \frac{15.9-10.6}{1.6}]$

$=P(-2.1875\leq Z\leq3.3125 )$

  $=P(3.3125)-P(-2.1875)$

  $=0.9996-0.0143$

$=0.9853$

d)$P(x\geq 5.3)=P [\frac{X-\mu}{\sigma }\geq \frac{5.3-10.6}{1.6}]$

  $=P [Z\geq-3.3125]$

  $=1-P(Z<-3.3125)$

$=1-0.0004$

Probability  $=0.9996$

  

e)$P(x\leq 13.4)=P[\frac{X-\mu}{\sigma }\leq \frac{13.4-10.6}{1.6}]$

$=P[Z\leq 1.75]$

  $=P(Z<1.75)$

Probability  $=0.9599$

Note [using standard Normal table.]