Question

A Food Marketing Institute found that 26% of households spend more than $125 a week on groceries. Assume the population proportion is 0.26 and a simple random sample of 159 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.27 and 0.49? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Do not use tables unless specifically told to. Answer = (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.26
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.26*0.74/159)
=0.0348

the probability that the sample proportion of households spending more than $125 a week is between 0.27 and 0.49
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.27) = (0.27-0.26)/0.0348
= 0.01/0.0348 = 0.2874
= P ( Z <0.2874) From Standard Normal Table
= 0.61308
P(X < 0.49) = (0.49-0.26)/0.0348
= 0.23/0.0348 = 6.6092
= P ( Z <6.6092) From Standard Normal Table
= 1
P(0.27 < X < 0.49) = 1-0.61308 = 0.3869