Question

The mean of a normal probability distribution is 200; the standard deviation is 10. Refer to the table in Appendix B.1. (Round the final answers to 2 decimal places.)   

a. About what percentage of the observations lie between 190 and 210?

Percentage of observations            %

b. About what percentage of the observations lie between 180 and 220?

   

Percentage of observations            %

c. About what percentage of the observations lie between 170 and 230?

   

Percentage of observations            %

Answer 1

Solution :

Given that ,

a.

P(190 < x < 210) = P[(190 - 200)/ 10) < (x - ) /  < (210 - 200) / 10) ]

= P(-1 < z < 1)

= P(z < 1) - P(z < -1)

= 0.8413 - 0.1587

= 0.6826

Percentage of observations 68.26%

b.

P(180 < x < 220) = P[(180 - 200)/ 10) < (x - ) /  < (220 - 200) / 10) ]

= P(-2 < z < 2)

= P(z < 2) - P(z < -2)

= 0.9772 - 0.0228

= 0.9544

Percentage of observations 95.44%

c.

P(170 < x < 230) = P[(170 - 200)/ 10) < (x - ) /  < (230 - 200) / 10) ]

= P(-3 < z < 3)

= P(z < 3) - P(z < -3)

= 0.9987 - 0.0013

= 0.9974

Percentage of observations 99.74%