Question

A Food Marketing Institute found that 34% of households spend more than $125 a week on groceries. Assume the population proportion is 0.34 and a simple random sample of 101 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is less than 0.35?

There is a  probability that the sample proportion of households spending more than $125 a week is less than 0.35. Round the answer to 4 decimal places. to find answer

Answer 1

Solution

Given that,

p = 0.34

1 - p = 1 - 0.34 = 0.66

n = 101

= p = 0.34

=  [p ( 1 - p ) / n] =   [(0.34 * 0.66) / 101 ] = 0.0471

P( < 0.35)

= P[( - ) / < (0.35 - 0.34) / 0.0471]

= P(z < 0.21)

Using z table,

= 0.5832