A Food Marketing Institute found that 42% of households spend more than $125 a week on groceries. Assume the population proportion is 0.42 and a simple random sample of 55 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.38 and 0.5?
Note: You should carefully round any z-values you calculate to 4
decimal places to match wamap's approach and calculations.
Answer = (Enter your answer as a number accurate to 4 decimal
places.)
Solution:
mean=p=0.42
standard deviation=sqrt(p*(1-p)/n=sqrt(0.42*(1-0.42)/55)=0.06655142
P(0.38<p^<0.5)
z=p^-p/sqrt(p*(1-p)/n)
P(0.38-.42/0.06655142<Z<0.5-0.42/0.06655142)
P(-0.601039<Z<1.202078)
P(-0.6010<Z<1.2021)
P(Z<1.2021)-P(Z<-0.6010)
=0.8853-0.2739
=0.6114
ANSWER:
0.6114
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