Question

e mean of a normal distribution is 540 kg. The standard deviation is 20 kg. Refer to the table in Appendix B.1. (Round the z values to 2 decimal places and the final answers to 4 decimal places.) a. What is the area between 547 kg and the mean of 540 kg? Area b. What is the area between the mean and 522 kg? Area c. What is the probability of selecting a value at random and discovering that it has a value of less than 522 kg? Probability

Answer 1

Solution :

Given that ,

mean = = 540

standard deviation = = 20

(a)

P( 540 < x < 547 ) = P(( 540-540 / 20 ) < (x - ) /  < (547 - 540) / 20) )

= P( 0 < z <  0.35 )

= P(z < 0 .35) - P(z < 0 )

= 0.6368 - 0.5

= 0.1368

Area = 0.1368

(b)

P( 522 < x < 540 ) = P(( 522 - 540 / 20 ) < (x - ) /  < (540 - 540) / 20) )

= P( - 0.9 < z <  0 )

= P(z < 0) - P (z < - 0.9 )

= 0.5 - 0.1841

= 0.3159

Area = 0.3159

(c)

P(x < 522 ) = P((x - ) / < ( 522- 540) / 20)

= P(z < -0.9 )

= 0.1841

Probability = 0.1841