Question

             Suppose that blood alcohol content of students who drink 5 beers varies from student to student according to a normal distribution with mean 0.08 and standard deviation 0.01.

___9. The middle 68% of students who drink 5 beers have blood alcohol content between about A. 0.07 and 0.09      B.  0.06 and 0.10      C.  0.05 and 0.11      D.  0.04 and 0.12

___ 10. What percent of students who drink 5 beers have blood alcohol content below 0.08 (the legal limit for               driving in most states)?

A.  2.5%      B.  5%      C.  16%      D.  32%     E.  50%

Answer 1

Solution:-

Given that,

mean = = 0.08

standard deviation = = 0.01

a) Using standard normal table,

P( -z < Z < z) = 68%

= P(Z < z) - P(Z <-z ) = 0.68

= 2P(Z < z) - 1 = 0.68

= 2P(Z < z) = 1 + 0.68

= P(Z < z) = 1.68 / 2

= P(Z < z) = 0.84

= P(Z < 0.99 ) = 0.84

= z  ± 0.99

Using z-score formula,

x = z * +

x = -0.99 * 0.01 + 0.08

x = 0.07

Using z-score formula,

x = z * +

x = 0.99 * 0.01 + 0.08

x = 0.09

The middle 68% of students who drink 5 beers have blood alcohol content between 0.07 and 0.09

correct option is =A

b) P(x < 0.08)

= P[(x - ) / < (0.08 - 0.08) /0.01 ]

= P(z < 0)

Using z table,

= 0.5

= 50%

correct option is = E