Question

The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Suppose that 43 randomly selected people are observed pouring syrup on their pancakes. Round all answers to 4 decimal places where possible.
What is the distribution of XX? XX ~ N(_______,_________)
What is the distribution of ¯xx¯? ¯xx¯ ~ N(______,_________)
If a single randomly selected individual is observed, find the probability that this person consumes is between 61.4 mL and 62.8 mL. ________
For the group of 43 pancake eaters, find the probability that the average amount of syrup is between 61.4 mL and 62.8 mL. _________
For part d), is the assumption that the distribution is normal necessary? No Yes
please only answer if you are able to answer all parts correctly

Answer 1

Answer:

(a) X ~ N($\mu=63, \sigma^{2} = 13^{2}$).

    $\bar X$ ~ N($\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = amount of syrup that people put on their pancakes

The z-score probability distribution for the normal distribution is given by;

                      Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean amount of syrup = 63 mL

            $\sigma$ = standard deviation = 13 mL

So, the distribution of X ~ N($\mu=63, \sigma^{2} = 13^{2}$).

Let $\bar X$ = sample mean amount of syrup that people put on their pancakes

The z-score probability distribution for the sample mean is given by;

                      Z  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\mu$ = mean amount of syrup = 63 mL

            $\sigma$ = standard deviation = 13 mL

            n = sample of people = 43

So, the distribution of $\bar X$ ~ N($\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

   P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X $\leq$ 61.4 mL)

  P(X < 62.8 mL) = P( $\frac{X-\mu}{\sigma}$ < $\frac{62.8-63}{13}$ ) = P(Z < -0.02) = 1 - P(Z $\leq$ 0.02)

                                                           = 1 - 0.50798 = 0.49202

  P(X $\leq$ 61.4 mL) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{61.4-63}{13}$ ) = P(Z $\leq$ -0.12) = 1 - P(Z < 0.12)

                                                           = 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < $\bar X$ < 62.8 mL)

   P(61.4 mL < $\bar X$ < 62.8 mL) = P($\bar X$ < 62.8 mL) - P($\bar X$ $\leq$ 61.4 mL)

  P($\bar X$ < 62.8 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{62.8-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z < -0.10) = 1 - P(Z $\leq$ 0.10)

                                                           = 1 - 0.53983 = 0.46017

  P($\bar X$ $\leq$ 61.4 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{61.4-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z $\leq$ -0.81) = 1 - P(Z < 0.81)

                                                           = 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.