Question

The amount of syrup that people put on their pancakes is normally distributed with mean 60 mL and standard deviation 11 mL. Suppose that 12 randomly selected people are observed pouring syrup on their pancakes. Round all answers to 4 decimal places where possible.
What is the distribution of
X
?
X
~ N(
,
)
What is the distribution of
¯
x
?
¯
x
~ N(
,
)
If a single randomly selected individual is observed, find the probability that this person consumes is between 59.3 mL and 61.2 mL.
For the group of 12 pancake eaters, find the probability that the average amount of syrup is between 59.3 mL and 61.2 mL.
For part d), is the assumption that the distribution is normal necessary? YesNo

Answer 1

Using the normal distribution and the central limit theorem, we have that:

• The distribution of X is $X \approx N(60,11)$.

• The distribution of $\bar{X}$ is $\bar{X} \approx (60,3.1754)$.

• 0.0637 = 6.37% probability that a single person consumes between 59.3 mL and 61.2 mL.

• 0.2351 = 23.51% probability that the sample mean of the consumption of 12 people is between 59.3 mL and 61.2 mL. Since the sample size is less than 30, a normal distribution has to be assumed.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

In this problem, the parameters are given as follows:

$\mu = 60, \sigma = 11, n = 12, s = \frac{11}{\sqrt{12}} = 3.1754$.

Hence:

• The distribution of X is $X \approx N(60,11)$.

• The distribution of $\bar{X}$ is $\bar{X} \approx (60,3.1754)$.

The probabilities are given by the p-value of Z when X = 61.2 subtracted by the p-value of Z when X = 59.3, hence, for a single individual:

X = 61.2:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{61.2 - 60}{11}$

Z = 0.11

Z = 0.11 has a p-value of 0.5398.

X = 59.3:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{59.3 - 60}{11}$

Z = -0.06

Z = -0.06 has a p-value of 0.4761.

0.5398 - 0.4761 = 0.0637.

0.0637 = 6.37% probability that a single person consumes between 59.3 mL and 61.2 mL.

For the sample of 12, we have that:

X = 61.2:

$Z = \frac{X - \mu}{s}$

$Z = \frac{61.2 - 60}{3.1754}$

Z = 0.38

Z = 0.38 has a p-value of 0.6480.

X = 59.3:

$Z = \frac{X - \mu}{s}$

$Z = \frac{59.3 - 60}{3.1754}$

Z = -0.22

Z = -0.22 has a p-value of 0.4129.

0.6480 - 0.4129 = 0.2351 = 23.51% probability that the sample mean of the consumption of 12 people is between 59.3 mL and 61.2 mL. Since the sample size is less than 30, a normal distribution has to be assumed.

More can be learned about the normal distribution and the central limit theorem at brainly.com/question/24188986

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