Answer to Question #238409 in Calculus for Arman

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 sample one of ten customers purchased different item from central store i.e. 11, 37, 25, 62, 51, 42, 23, 53,17, 18, 19 11, 32, 20 and the sample of other ten customers purchased different items from central store i.e 5,11,14,7,18,12, 13, 14, 15 16,18,17,16,11. If a random sample of ten items are selected of both customers, what is the probability that customer will have a mean item which is a) at least thirty one b) at most fourty five c) between thirty nine and fifty six d) mean item is fifty percent and e) mean item is hundred percent.

Answer 1

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)

Ordinal Variables- is a type of measurement variable that accepts values in a particular order or rank in this case we can group age, number of children per family, length of hair, the height of a book, number of citizens of a country as ordinal variables.

A nominal variable is a variable that is used to name, label or categorize specific attributes being measured. It accepts qualitative values representing various categories like a watch, interest, hate, love, the color of shirts, and willing.

The interval variable is a measurement variable that is used to define values measured along a scale, with each point being placed at an equal distance from the next. For example life of a person, speedometer, temperature, and life of a bulb.

Ratio Variable- the height of a tree, employee performance this is so because all this when measured they can include a zero value.