Question

The average price of personal computers manufactured by MNM Company is $1,200 with a standard deviation of $220. Furthermore, it is known that the computer prices manufactured by MNM are normally distributed.

a. What is the probability that a randomly selected computer will have a price of at least $890?  

b. Computers with prices of more than $1,570 receive a discount. What percentage of the computers will receive the discount?

What is the minimum value of the middle 95% of MNM computer prices? the minimum =$

Answer 1

Solution :

a)

Given that ,

mean =

standard deviation =

P(x > $890)= 1 - P(x<890)

(B)

P(x >$1570 ) = 1 - P(x<1570 )

Using z table

= 1 - 0.95352

= 0.04648

answer=4.68%

(C)

middle 95% of score is

P(-z < Z < z) = 0.95

P(Z < z) - P(Z < -z) = 0.95

2 P(Z < z) - 1 = 0.95

2 P(Z < z) = 1 + 0.95 = 1.95

P(Z < z) = 1.95 / 2 = 0.975

P(Z <1.96 ) = 0.975

z ± 1.96 (see the probability 0.975 in standard normal (Z) table corresponding z value is 1.96 )

Using z-score formula

x= -1.96 *220+1200

x= 768.80

z = 1.96

Using z-score formula

x= 1.96 *220+1200

x= 1631.20

answer = Minimum Answer 768.80 , Maximum Answer 1631.20

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