A Food Marketing Institute found that 30% of households spend more than $125 a week on groceries. Assume the population proportion is 0.3 and a simple random sample of 245 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 0.32

75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For the sampling distribution of a proportion p with size n, we have that

In this problem, we have that:

So

What is the probability that the sample proportion of households spending more than $125 a week is less than 0.32

This is the pvalue of Z when X = 0.32. So

has a pvalue of 0.7517

75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32

p> q if you calculate p is100 q is 79

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.