We are asked to calculate the probability of having 10 or fewer households using the product in a sample of 40 households, assuming that the company's claim is correct.

Given Information:

• The company's claim is that 50% of households use the product, so the probability of success (household using the product) is $p = 0.50$.
• The sample size is $n = 40$ households.
• We are asked to find the probability that 10 or fewer households out of 40 use the product.

Since the problem involves a binomial distribution and asks to approximate with a normal distribution, we will first determine the mean and standard deviation of the binomial distribution, and then approximate it with a normal distribution.

Step 1: Define the Binomial Distribution

A binomial distribution $X$ can be approximated as a normal distribution when the sample size $n$ is large enough. For a binomial distribution:

• Mean: $\mu = np$
• Standard deviation: $\sigma = \sqrt{np(1 - p)}$

Here:

• $n = 40$
• $p = 0.50$

Thus: $\mu = 40 \times 0.50 = 20$ $\sigma = \sqrt{40 \times 0.50 \times (1 - 0.50)} = \sqrt{40 \times 0.50 \times 0.50} = \sqrt{10} \approx 3.162$

Step 2: Normal Approximation

To use the normal distribution, we use the continuity correction since we are approximating a discrete binomial distribution with a continuous normal distribution.

We want the probability of having 10 or fewer successes, so we need to calculate:

$P(X \leq 10) \approx P\left( Z \leq \frac{10 + 0.5 - \mu}{\sigma} \right)$

where:

• $Z$ is the standard normal variable.
• $10 + 0.5$ is the continuity correction (we add 0.5 because we are approximating a discrete value with a continuous distribution).

Step 3: Calculate the Z-Score

Substitute the values into the formula for the Z-score:

$Z = \frac{10.5 - 20}{3.162} = \frac{-9.5}{3.162} \approx -3.00$

Step 4: Find the Probability

Now, we use the standard normal distribution table or a calculator to find the probability corresponding to a Z-score of $-3.00$.

Using the standard normal distribution table: $P(Z \leq -3.00) \approx 0.00135$

Thus, the probability of having 10 or fewer households using the product in a sample of 40 is approximately 0.00135, or 0.135%.

Conclusion:

The probability of observing 10 or fewer households using the product, given the company's claim, is very low. This suggests that the company's claim of 50% usage might not be accurate, based on this sample. The observed sample of 10 out of 40 households could be considered unusual if the claim were true.

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Would you like more details on any step of the solution? Here are 5 related questions to explore further:

1. How do you determine when it's appropriate to approximate a binomial distribution with a normal distribution?
2. What are the assumptions behind using the normal approximation for a binomial distribution?
3. How does the continuity correction affect the accuracy of the normal approximation?
4. What other methods can we use to calculate probabilities for binomial distributions if normal approximation isn't suitable?
5. How do you interpret the result of a hypothesis test comparing observed data to a claimed proportion?

Tip: When approximating a binomial distribution with a normal distribution, ensure that $np$ and $n(1-p)$ are both greater than 5. This is a common rule of thumb for ensuring the approximation is accurate.