[Probablity and Statistics] Sampling and Sampling Distribution ~ Đặng Pete Tutoring

Thursday, June 29, 2023

[Probablity and Statistics] Sampling and Sampling Distribution

I. Sample Distribution of Sample Mean

Prob1. Assume the prices of meals at a restaurant are normally distributed with an average of $35 and a standard deviation of $5. I have $33 in my pocket.

1. What is the probability I have enough money for my meal ?

2. I’m out to dinner with friends and we’re splitting the check evenly by 10.

a. What is the expected price per meal now ?

b. What is the probability I have enough moneny now ?

c. Why did the probability change ?

Solution:

$1.{\rm{ }}P(X \le 33) = P(Z \le \frac{{33 - 35}}{5}) = P(Z \le - 0.4) = 0.3446$

$2a.{\rm{ }}E(\overline X ) = \mu = 35\$ ;{\rm{ }}\overline X \sim N(\mu ,\frac{\sigma }{{\sqrt n }})$

$2b.{\rm{ }}P(\overline X \le 33) = P(Z \le \frac{{33 - 35}}{{(5/\sqrt {10} }}) = P(Z \le - 1.26) = 0.1038$

$2c.{\rm{ }}We're{\rm{ }}averaging{\rm{ }}10{\rm{ }}values,{\rm{ }}the{\rm{ }}std{\rm{ }}errror{\rm{ }}is{\rm{ }}less{\rm{ }}and{\rm{ }}values{\rm{ }}convergence{\rm{ }}toward{\rm{ }}\$ 35$

Prob2. You manage a fast-food franchise in LA. Last month the mean waiting time at the drive-through windown for branches in LA was 3.8 minutes and the standard deviation was 0.95 minutes. You can assume the waiting times are normally distributed.

a. Let X denote the waiting time of a random customer. What is the probability that X is between 3 and 5 minutes ?

b. What is the probability the average waiting time of the next 10 customers is greater than 3 minutes ?

c. Suppose your restaurant has 10 customers in line for picking up lunch. What is the total amount of time required so that with probability 90% that all 10 orders can be completed during that period ?

Solution:

$a.{\rm{ }}X \sim N(\mu ,\sigma );{\rm{ }}\mu = 3.8;{\rm{ }}\sigma = 0.95$

$P(3 \le X \le 5) = P(X \le 5) - P(X \le 3) = P(Z \le \frac{{5 - 3.8}}{{0.95}}) - P(Z \le \frac{{3 - 3.8}}{{0.95}}) = ?$

$b.{\rm{ }}\overline X \sim N({\mu _{\overline x }},{\sigma _{\overline x }});$ ${\rm{ }}{\mu _{\overline x }} = \mu = 3.8;{\rm{ }}{\sigma _{\overline x }} = \frac{\sigma }{{\sqrt n }} = \frac{{0.95}}{{\sqrt {10} }}$

${\rm{ }}P(\overline X > 3) = P(Z > \frac{{3 - 3.8}}{{(0.95/\sqrt {10} }}) = P(Z > ?) = 1 - P(Z < ?) = ?$

$c.{\rm{ }}P(\overline X \le t) = 0.9 \Rightarrow P(\overline X \le t) = P(Z \le \frac{{t - 3.8}}{{(0.95/\sqrt {10} )}}) = P(Z \le \frac{{t - 3.8}}{{0.3}})$

$\Rightarrow P(Z \le \frac{{t - 3.8}}{{0.3}}) = 0.9 = P(Z \le 1.28) \Rightarrow \frac{{t - 3.8}}{{0.3}} = 1.28 \Rightarrow t = 4.184$

II. Sample Distribution of Sample Proportion

Prob3. A study found that 55% of British firms experienced a cyber-attack in the past year.

a. What are the expected value and the standard error of the sample proportion derived from a random sample of 100 firms ?

b. What are the expected value and the standard error of the sample proportion derived from a random sample of 200 firms ?

c. Comment on the value of the standard error as the sample size gets larger.

Solution:

$\begin{array}{l} a.{\rm{ }}p = 0.55;n = 100;E(\overline p ) = p = 0.55\\ se(\overline p ) = \sqrt {\frac{{p(1 - p)}}{n}} = \sqrt {\frac{{0.55(1 - 0.55)}}{{100}}} = 0.0497 \end{array}$

$\begin{array}{l} b.{\rm{ }}p = 0.55;n = 200;E(\overline p ) = p = 0.55\\ se(\overline p ) = \sqrt {\frac{{p(1 - p)}}{n}} = \sqrt {\frac{{0.55(1 - 0.55)}}{{200}}} = 0.0352 \end{array}$