Answer :
To find the probability that the mean weight of a sample of 25 potatoes is between 7 and 8 ounces, we need to use the properties of the normal distribution. Given that the weights of the potatoes are normally distributed with a mean ([tex]\mu[/tex]) of 7.4 ounces and a standard deviation ([tex]\sigma[/tex]) of 2.2 ounces, we can start by considering the distribution of the sample mean.
The mean of the sampling distribution of the sample mean ([tex]\bar{x}[/tex]) for a sample size [tex]n = 25[/tex] remains [tex]\mu = 7.4[/tex] ounces. However, the standard deviation of the sample mean, often called the standard error, is given by the formula:
[tex]\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{2.2}{\sqrt{25}} = \frac{2.2}{5} = 0.44[/tex]
Now, we want to find the probability that the sample mean is between 7 and 8 ounces. We first convert these weights to their corresponding [tex]z[/tex]-scores using the formula:
[tex]z = \frac{\bar{x} - \mu}{\text{SE}}[/tex]
For [tex]\bar{x} = 7[/tex]:
[tex]z = \frac{7 - 7.4}{0.44} = \frac{-0.4}{0.44} \approx -0.91[/tex]For [tex]\bar{x} = 8[/tex]:
[tex]z = \frac{8 - 7.4}{0.44} = \frac{0.6}{0.44} \approx 1.36[/tex]
Next, we find the probabilities corresponding to these [tex]z[/tex]-scores using a standard normal distribution table or a calculator:
- [tex]P(Z < -0.91) \approx 0.1814[/tex]
- [tex]P(Z < 1.36) \approx 0.9131[/tex]
Thus, the probability that the sample mean is between 7 and 8 ounces is:
[tex]P(-0.91 < Z < 1.36) = P(Z < 1.36) - P(Z < -0.91)[/tex]
[tex]= 0.9131 - 0.1814 = 0.7317[/tex]
Therefore, there is approximately a 73.17% probability that the mean weight of the sample of 25 potatoes is between 7 and 8 ounces.