A Food Marketing Institute found that 31% of households spend more than $125 a week on groceries. Assume the population proportion is 0.31 and a simple random sample of 373 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 0.33

0.7967

Step-by-step explanation:

We know that population proportion p=0.31.

We have to find P(phat<0.33).

Mean=p=0.31

standard deviation=0.024 (rounded to three decimal places)

Thus, the required probability that sample proportion of households spending more than $125 a week is less than 0.33 is 79.67%

answer: 48

just remember to use pemdas:

parentheses

expressions

multiplication

division

addition

subtraction

step-by-step explanation:

the true ones are a b and d