The distribution of syrup amounts follows a normal distribution, and for a sample mean of 42 people, the assumption of normality is valid, allowing us to calculate probabilities accurately.
a. The distribution of X for a single observation of syrup poured on pancakes follows a normal distribution with a mean of 64 mL and a standard deviation of 9 mL. So,

b. The distribution of
the sample mean for a group of 42 individuals, also follows a normal distribution.
For the sample mean, the mean
remains the same as the population mean (64 mL), and the standard deviation
is calculated as
is the population standard deviation and n is the sample size.
Therefore,

c. To find the probability that a single randomly selected individual consumes between 65.9 mL and 67 mL, we'll use the Z-score formula and then find the probability using the standard normal distribution:
![\[Z = (X - \mu)/(\sigma)\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/ukar93hqaj7c9dkxgqae4auvhjua7sx0pg.png)
Where X is the value (67 mL),
is the mean (64 mL), and
is the standard deviation (9 mL).
![\[Z_(67) = (67 - 64)/(9) = (3)/(9) = 0.3333\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/2btrroc74wp9pb4xzfqub8a6i124la1dn6.png)
Now, we find the corresponding probabilities using a standard normal distribution table or calculator.
![\[P(65.9 < X < 67) = P(0 < Z < 0.3333)\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/uejopo0fjehgrins3mllm77ubc0gj77mo2.png)
d. For the group of 42 pancake eaters, finding the probability that the average amount of syrup is between 65.9 mL and 67 mL involves using the properties of the sample mean distribution:
![\[P(65.9 < \bar{X} < 67) = P\left(\frac{65.9 - \mu_{\bar{X}}}{\sigma_{\bar{X}}} < Z < \frac{67 - \mu_{\bar{X}}}{\sigma_{\bar{X}}}\right)\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/n8mfloboazs5zfzuq43ocshq0ij78akwox.png)
Substitute the values of
into the formula and use the Z-score approach similar to part c to find this probability.
e. For part d), the assumption that the distribution is normal is necessary due to the central limit theorem, which states that the sample mean of a sufficiently large sample size from any population will be approximately normally distributed. Therefore, with a sample size of 42 (which is considered relatively large), the normality assumption holds for the sample mean distribution.