Question

A lab orders 100 rats a week. Prices for 100 rats follow the following​ distribution:

Price (x) in $	Probability (P)	X^2	X*P	X^2*P
15	0.39
20	0.34
25	0.27

A. The probability that the price of purchasing 100 rats is​ $20.0 is:

​B. The probability that the price of purchasing 100 rats is less than​ $21.0 is:

C. The expected price of ordering 100 rats per week is $:

D. The variance for the price of ordering 100 rats per week is:

E. The standard deviation for the price of ordering 100 rats per week is:

Answer 1

solution:

From the given information

A lab orders 100 rats a week. Prices for 100 rats follow the following​ distribution:   

Price (x) in$	Probability P(X)	X^2	X*P(X)	X^2 * P(X)
15	0.39	225	5.85	87.75
20	0.34	400	6.80	136
25	0.27	625	6.75	168.75
		Total	19.4	392.5

A) Probability that price of purchasing 100 rats is $20 is = P(X=20) = 0.34

B)  The probability that the price of purchasing 100 rats is less than​ $21.0 = P(X<21)

= P(X=15) + P(X=20)

= 0.39 + 0.34

= 0.73

C) The expected price of ordering 100 rats per week = E{X] = = $19.4

X* P(X)

D) The variance for the price of ordering 100 rats per week ($\sigma^2$) = E[X^2]-E[X]^2

= $\sum X^2*P(X)$ - ()^2

X* P(X)

= 392.5 - (19.4)^2

= $16.14

E) The standard deviation for the price of ordering 100 rats per week ($\sigma$) = $\sqrt 16.14$ =~ $ 4.017