Question

n=600

p=.29

range is 15000 to 29999

using normal distribution approximation with continuity correction round z scores to 2 decimal places before looking them up or calculating four decimal places in final answer

a) exactly 155

b) between 184 and 196 inclusive

c) no more than 150

can this problem be done on ti 84 plus ce

I did it on paper and got -1 for a?? so obviously I am not doing them correctly

Answer 1

n=600

p=.29

range is 15000 to 29999

using normal distribution approximation with continuity correction round z scores to 2 decimal places before looking them up or calculating four decimal places in final answer

Expectation = np = 174

Variance = np(1 - p) = 123.54

Standard deviation = 11.1149

1. exactly 155

This is between 154.5 to 155.5 with continuity correction.

Z value for 154.5, z =(154.5-174)/11.1149 = -1.75

Z value for 155.5, z =(155.5-174)/11.1149 = -1.66

P( x=155)= P( -1.75<z<-1.66) = P( z <-1.66)-P( z < -1.75)

=0.0485-0.0401

=0.0084

b) between 184 and 196 inclusive

Z value for 183.5, z =(183.5-174)/11.1149 = 0.85

Z value for 196.5, z =(196.5-174)/11.1149 = 2.02

P( 184≤ x≤196) = P( 0.85<z<2.02) = P( z <2.02)-P( z <0.85)

=0.9783 -0.8023

=0.1760

2. no more than 150

we have to find P( x ≤150)

Z value for 150.5, z =(150.5-174)/11.1149 = -2.11

P( x ≤150) = P( z < -2.11)

= 0.0174