Question

A survey found that the mean amount customers spend for a valentine's gift for their pets is $5.05. Assume the distribution of the amount spent is approximately normal and the standard deviation is $0.56. If a random sample of 47 pet owners is selected, find the probability that the mean amount spent of the sample will be less than $5.30?

Answer 1

I'd be glad to help!<3

To find the probability that the mean amount spent of the sample is less than $5.30, we need to standardize the sample mean and use the standard normal distribution.

Let's define the following variables:
\( \mu \) = population mean (mean amount customers spend for a valentine's gift for their pets), which is \$5.05
\( \sigma \) = population standard deviation, which is \$0.56
\( n \) = sample size, which is 47
\( \overline{x} \) = sample mean amount spent

Now, we need to calculate the z-score using the formula:
\[ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

Substituting the given values:
\[ z = \frac{5.30 - 5.05}{\frac{0.56}{\sqrt{47}}} \]

Calculating the value inside the square root:
\[ \frac{0.56}{\sqrt{47}} = 0.0818 \]

Substituting this value:
\[ z = \frac{5.30 - 5.05}{0.0818} \]

Calculating the numerator:
\[ 5.30 - 5.05 = 0.25 \]

Substituting this value:
\[ z = \frac{0.25}{0.0818} \]

Calculating the value:
\[ z \approx 3.05 \]

Now, we need to find the probability that the standardized sample mean is less than \( z = 3.05 \). This can be done by looking up the z-score in the standard normal distribution table. However, since the z-score is larger than 3, it is so close to the mean, the probability is nearly 1 (as the z-score approaches infinity, the probability approaches 1).

So, the probability that the mean amount spent of the sample is less than $5.30 is approximately 1.

Hope this helped!

Answer 2

To calculate the probability of the sample mean being less than $5.30, calculate the standard error, determine the z-score for $5.30, and then look up the corresponding probability in the standard normal distribution table.

Step-by-step explanation:

To find the probability that the mean amount spent of the sample will be less than $5.30 when the population mean is $5.05 and the population standard deviation is $0.56, we use the Central Limit Theorem since the sample size is sufficiently large (n=47). We start by calculating the standard error (SE) of the mean since we're dealing with a sample:

SE = σ / √n = $0.56 / √47

Next, we determine the z-score for $5.30 using the following formula:

Z = (X - μ) / SE = ($5.30 - $5.05) / SE

After calculating the Z-value, we refer to the standard normal distribution table to find the corresponding probability for the Z-value. This probability will give us the chance that the sample mean is less than $5.30.