Answer to Question #198925 in Statistics and Probability for Waqas

Question

A marketing research analyst selects a random sample of ten customers out of the fourty

who purchased a particular item from central store. The ten customers purchased different item 

from central store i.e. 11, 37, 25, 62, 51, 21, 18, 41, 32, 20. If a random sample of ten items are 

selected, what is the probability that customer will have a mean item which is a) at least thirty 

one b) at most fifty five c) between thirty nine and fifty six d) mean item is fifty percent and e) 

mean item is hundred percent.

b) ) From the following variables, Categories them and also write a reason? 

Number of children per family, Watch, Age, length of hair, Willing, weight of a box, love, height of 

a tree, color of Shirts, life of a person, interest, hate, Speedo meter, number of citizens of a 

country, employee performance, behavior, temperature, life of bulb.

Answer 1

A) Data : 11, 37, 25, 62, 51, 21, 18, 41, 32, 20

Mean $\mu=\dfrac{\sum x}{n}=\dfrac{318}{10}=31.8$

Standard Deviation $\sigma =\sqrt{\dfrac{\sum x^2-\frac{(\sum x)^2}{n}}{n}}=\sqrt{\dfrac{12430-\frac{(318)^2}{10}}{10}}$

$=\sqrt{\dfrac{2317.6}{10}}=\sqrt{231.76}=15.22$

a) at least thirty one

$X=31\\\mu=31.8\\\sigma=15.22$

Using normal distribution

$P(X\geq 31)$

$z=\dfrac{X-\mu}{\sigma/\sqrt n}=\dfrac{31-31.8}{15.22/\sqrt{10}}=\dfrac{-0.8\times 3.16}{15.22}=-0.1662$

Using Standard Normal Distribution Table:

$P(X\geq31)=P(z\geq -0.1662)=0.434$

b)  at most fifty five

$X=55\\\mu=31.8\\\sigma=15.22$

$P(X\leq55)$

$z=\dfrac{X-\mu}{\sigma/\sqrt n}=\dfrac{55-31.8}{15.22/\sqrt{10}}=4.820$

Using Standard Normal Distribution Table:

$P(X\leq 55)=P(z\leq 4.820)\approx1$

c) between thirty nine and fifty six

$X_1=39\\X_2=56\\\mu=31.8\\\sigma= 15.22$

$P(39\leq X\leq 56)$

$z_1=\dfrac{X_1-\,u}{\sigma/\sqrt n}=\dfrac{39-31.8}{15.22/\sqrt{10}}=1.4959\\\ \\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt n}=\dfrac{56-31.8}{15.22/\sqrt{10}}=5.028$

Using Standard Normal Distribution Table:

$P(39\leq X\leq 56)=P(1.4959\leq z\leq 5.028)=0.0673$

d)  mean item is fifty percent

$X=50\\\mu=31.8\\\sigma=15.22$

$P(X\leq50)$

$z=\dfrac{X-\mu}{\sigma/\sqrt n}=\dfrac{50-31.8}{15.22/\sqrt{10}}=3.7814$

Using Standard Normal Distribution Table:

$P(X\leq50)=P(z\leq3.7814)\approx 0.9999$

e)  mean item is hundred percent

$X=100\\\mu=31.8\\\sigma=15.22$

$P(X\leq100)$

$z=\dfrac{X-\mu}{\sigma/\sqrt n}=\dfrac{100-31.8}{15.22/\sqrt{10}}=14.16$

Using Standard Normal Distribution Table:

$P(X\leq 100)=P(z\leq 14.16)\approx 1$

B) 1) First category:

Number of children per family, Age, Love, The life of a person, Interest, Hate

This data is selected because this will show the social life of a citizen in a country.

2 ) Second category:

Employee Performance, Behavior, Length of hair, Willing

This data is selected to show the work pressure in an office and stress level of employees.

3) Third category:

Weight of a box, The color of shirts, The life of bulb, Watch

This data will show the type of things in a company and the weight of consignment accordingly.

4) Fourth category:

Height of a tree, Number of citizens of a country, Speedometer, Temperature

This data will describe the impact by different countries on global warming.