Answer:
0.6210
Step-by-step explanation:
Given that a Food Marketing Institute found that 39% of households spend more than $125 a week on groceries
Sample size n =87
Sample proportion will follow a normal distribution with p =0.39
and standard error = [tex]\sqrt{\frac{0.39(1-0.39)}{87} } \\=0.0523[/tex]
the probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41
=[tex]P(0.29<p<0.41)\\\\ =0.64892-0.02794\\=0.6210[/tex]
There is 0.6210 probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41
