n=600 p=.29 range is 15000 to
n=600
p=.29
range is 15000 to 29999
using normal distribution approximation with continuitycorrection round z scores to 2 decimal places before looking themup 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 doingthem correctly
Answer:
n=600
p=.29
range is 15000 to 29999
using normal distribution approximation with continuitycorrection round z scores to 2 decimal places before looking themup or calculating four decimal places in final answer
Expectation = np = 174
Variance = np(1 – p) = 123.54
Standard deviation = 11.1149
- exactly 155
This is between 154.5 to 155.5 withcontinuity 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
- 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
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