(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:
$ X ~ N(\mu = 60, \sigma = 12) $
B) The distribution of $ x\bar $ is normal distribution:
$ x\bar ~ N(\mu_(x\bar), \sigma_(x\bar)) $
→ The mean of sampling distribution $ (\mu_(x\bar)) : $
$ \mu_(X\bar) = \mu = 60 $
→ The standard deviation of sampling distribution $ (\sigma_(x\bar)) : $
$ \sigma_(X\bar) = \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\left(\frac{61.5 – 60}{12} < z < \frac{63.5 – 60}{12}\right) $
$ = 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 < \bar{x} < 63.5) = P\left(\frac{61.5 – 60}{\frac{12}{\sqrt{50}}} < z < \frac{63.5 – 60}{\frac{12}{\sqrt{50}}}\right) $
$ = P(0.884 < z < 2.062) $
$ = P(z < 2.062) – P(z < 0.884) = 0.9804 – 0.8117 $
$ = 0.1687 $
E) Yes
