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Let's solve the problem step-by-step.

### Given Data:
- Population proportion $ p = 0.35 $
- Sample size $ n = 281 $
- We are asked to find the probability that the sample proportion $ \hat{p} $ is less than 0.32.

### Step 1: Standard Error of the Sample Proportion
The standard error (SE) for the sample proportion is given by:

\[
SE = \sqrt{\frac{p(1 - p)}{n}}
\]

Substitute the values of $ p $ and $ n $:

\[
SE = \sqrt{\frac{0.35(1 - 0.35)}{281}}
\]

### Step 2: Z-Score Calculation
To find the probability, we first calculate the Z-score. The formula for the Z-score is:

\[
Z = \frac{\hat{p} - p}{SE}
\]

Where $ \hat{p} = 0.32 $ is the sample proportion. Substituting the values:

\[
Z = \frac{0.32 - 0.35}{SE}
\]

### Step 3: Finding the Probability
Once we compute the Z-score, we can use the standard normal distribution table to find the probability corresponding to that Z-score.

Let me calculate this step for you now.

A Food Marketing Institute found that 35% of households spend more than $125 a week on groceries. Assume the population proportion is 0.35 and a simple random sample of 281 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? Round the answer to 4 decimal places.

Find the probability that the sample proportion of households spending more than $125 a week on groceries is less than 0.32, given a population proportion of 0.35 and a sample size of 281 households. This problem involves calculating the standard error, Z-score, and using the normal distribution.

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