Suppose that the Blood Alcohol Content (BAC) of students who drink five beers varies from student to student according to a Normal distribution with mean 0.07 ans standard deviation 0.01. 1. The middle 95% of students who drink five beers have a BAC between

a. 0.06 and 0.08 b. 0.05 and 0.09 c. 0.04 and 0.10 d. 0.03 and 0.11

2. What percent of students who drink five beers have a BAC above 0.08 (the legal limit for driving in most states)?

a. 0.15% b. 0.3% c. 2.5% d. 16% e. 32%

3. What percent of students who drink five beers have a BAC above 0.10 (the legal limit for driving in most states)?

a. 0.15% b. 0.3% c. 2.5% d. 16% e. 32%

1.    b. 0.05 and 0.09

2.   d. 16%

3.    a. 0.15%

Step-by-step explanation:

Given that :

mean = 0.07

standard deviation = 0.01

Confidence interval = 95%

The level of significance ∝= 1 - 0.95 = 0.05

At 0.05 level of significance,

critical value for

critical value for  = 1.96

Confidence interval =

Lower limit =

Upper Limit =

Lower limit =

Upper limit =

Lower limit = 0.07 - 0.0196

Upper limit = 0.07 + 0.0196

Lower limit = 0.0504   0.05

Upper limit = 0.0896     0.09

The confidence interval of 95% is ( 0.05, 0.09)

2. What percent of students who drink five beers have a BAC above 0.08 (the legal limit for driving in most states)?

P(X > 0.08) = 1 - 0.8413

P(X > 0.08) = 0.1587

P(X > 0.08) 16%  

3. What percent of students who drink five beers have a BAC above 0.10 (the legal limit for driving in most states)?

P(X > 0.10) = 1 - 0.9987

P(X > 0.08) = 0.0013

P(X > 0.08) 0.15%  which is the closet value to 0.0013

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step-by-step explanation: ; )