The amount of syrup that people put on their pancakes is normally distributed with a mean of 60 mL and standard deviation of 12 mL. Suppose that 50 randomly selected people are observed pouring syrup on their pancakes. Round all answers to 4 decimal places where possible.

(a) What is the distribution of X? X ~ N(,)
(b) What is the distribution of x̄? x̄ ~ N(,)
(c) If a single randomly selected individual is observed, find the probability that the amount of syrup this person consumes is between 61.5 mL and 63.5 mL.
(d) For the group of 50 pancake eaters, find the probability that the average amount of syrup is between 61.5 mL and 63.5 mL.
(e) For part d), is the assumption that the distribution is normal necessary?

Answer:

Given Data:
The Population Mean $(\mu )=60$
The Population Standard Deviation $(\sigma )=12$
The Sample Size $(n)=50$

Solution:

A) The distribution of X is normal distribution:
$XN(\mu =60,\sigma =12)$

B) The distribution of $\text{Missing argument for bar}$ is normal distribution:
$x\overline{}N({\mu }_{(}x\overline{)},{\sigma }_{(}x\overline{)})$

→ The mean of sampling distribution $({\mu }_{(}x\overline{)}):$
${\mu }_{(}X\overline{)}=\mu =60$

→ The standard deviation of sampling distribution $({\sigma }_{(}x\overline{)}):$
${\sigma }_{(}X\overline{)}=\frac{\sigma }{\sqrt{n}}=\frac{12}{\sqrt{50}}=1.6971$

C) The probability that the amount of syrup this person consumes is between 61.5 mL and 63.5 mL:
$P(61.5<x<63.5)=P(\frac{61.5–60}{12}<z<\frac{63.5–60}{12})$
$=P(0.125<z<0.2917)$
$=P(z<0.2917)–P(z<0.125)=0.6146–0.5497$
$=0.0649$

D) For the group of 50 pancake eaters, the probability that the average amount of syrup is between 61.5 mL and 63.5 mL:
$P(61.5<\overline{x}<63.5)=P(\frac{61.5–60}{\frac{12}{\sqrt{50}}}<z<\frac{63.5–60}{\frac{12}{\sqrt{50}}})$
$=P(0.884<z<2.062)$
$=P(z<2.062)–P(z<0.884)=0.9804–0.8117$
$=0.1687$

E) Yes