Question

Please be thorough with the steps. The answers are in red.

Consider the non -standard normal random variable Zua , where 5.5 and a 1.9 μ.σ ι P (ZJ.g<-5.69) P(Z<5.69) =0.4602 P (Zua> 3.98) P (Z u.O 3.98) = 0.2119 P (Z 5.69,-3.98) P (ZL.aE [5.69, -3.98]) 0.3280 4,0 If P (Zo E(0o, a) 0.65 then a a = -4.7679 The amount of time a child in a certain region of Seoul, Korea, spends online on any given day is modeled (in hours) by the normal random variable Zuwith 3.8 and a= 1 What is the probability that the amount of time a randomly selected child from this region will spend online differs from by less than 0.6 hours? P (Zua-3.8 0.6) CHECK P (Z3.8 0.6)= 0.4515

Answer 1

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = -5.5

S.d = 1.9

1)

P(x<-5.69)

Z = (-5.69 - (-5.5))/1.9 = -0.1

From z table, P(z<-0.1) = 0.4602

2)

P(x>-3.98)

Z = (-3.98 - (-5.5))/1.9 = 0.8

From z table, P(z>0.8) = 0.2119

3)

P(-5.69<x<-3.98) = p(x<-3.98) - p(x<-5.69)

P(x<-3.98)

Z = 0.8

From z table, p(z<0.8) = 0.7881

P(x<-5.69) = 0.4602

Required answer is 0.7881 - 0.4602 = 0.3280

4)

From z table, p(z<0.39) = 0.65

So, 0.39 = (a - (-5.5))/1.9

A = -4.7679

5)

Here we need to find

P(x-0.6 < u <x+0.6)

As s.d = 1

Z = 0.6

So answer would be

P(z<0.6) - p(z<-0.6) = 0.4515