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

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khansawaqar7 khansawaqar7 · Answer 1 · 2020-08-16T15:56:49+02:00

Answer:

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

[tex]Standard deviation=\sqrt{\frac{p(1-p)}{n} }[/tex]

[tex]Standard deviation=\sqrt{\frac{0.31(0.69)}{373} }[/tex]

standard deviation=0.024 (rounded to three decimal places)

[tex]P(phat<0.33)=P(Z<\frac{0.33-0.31}{0.024})[/tex]

[tex]P(phat<0.33)=P(Z<\frac{0.02}{0.024})[/tex]

[tex]P(phat<0.33)=P(Z<0.83)[/tex]

[tex]P(phat<0.33)=0.5+0.2967[/tex]

[tex]P(phat<0.33)=0.7967[/tex]

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