I need help with statistics

Question

A food marketing Institute found that 33% of households spend more than $125 a week on groceries assume the population proportion is 0.33 and a simple random sample of 412 households is selected from the population, what is the probability that the sample portion of household spending more than $125 a week is less than 0.31?

there is a blank probability that the sample portion of household spending more than $125 a week is less than 0.31 round answer to four decimal places

Answer 1

33% of households spend more than $125 per week on groceries

population proportion = .33 =33%/100%= 33/100 = .33

what is the probability a random sample of 412 households spending more than $125 per week, will have a proportion less than .31?

calculate the z score, z= (x-mean)/s/sqrN

but s is not expressly given

to find s, use the binomial distribution, as there are two possibilities, either the household spends more than 125 or doesn't spend more than 125

mean = np = .33(412) = 135.96

standard deviation = square root of npq = np(1-q) = sqr(135.96(.67)) = sqr91.0932= about 9.5443

s= 9.5443

x= .31(412) = 127.72

z = (127.72- 135.96)/9.5443/sqr412

= -8.24sqr412/9.5443

= - 17.5239

find the corresponding area under the normal curve= nearly 0 % =0.0000% rounded to 4 decimals

or possibly the problem really wanted

z= -8.24/9.5443 = -0.8633

with corresponding area under the curve

= .19399

= 19.399% of the 412 households spent less than $125

if your computer isn't taking either of those answers, try 1-those answers

or 0.0000% or 80.6010%, although those last digits 990 or 010 may be off by that 3rd end digit

the "more" and "less" than in the problem gets confusing