Answer:
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 [tex]z_{\alpha/2} = z_{0.05/2}[/tex]
critical value for [tex]z_{0.025}[/tex] = 1.96
Confidence interval = [tex]\mathtt{\mu \pm ( {z} \times{\sigma})}[/tex]
Lower limit = [tex]\mathtt{\mu -( {z} \times{\sigma})}[/tex]
Upper Limit = [tex]\mathtt{\mu +( {z} \times{\sigma})}[/tex]
Lower limit = [tex]\mathtt{0.07 - ({1.96} \times {0.01})}[/tex]
Upper limit = [tex]\mathtt{0.07 + ({1.96} \times {0.01})}[/tex]
Lower limit = 0.07 - 0.0196
Upper limit = 0.07 + 0.0196
Lower limit = 0.0504 [tex]\simeq[/tex] 0.05
Upper limit = 0.0896 [tex]\simeq[/tex] 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)?
[tex]P(X> 0.08) = P(\dfrac{0.08 - \mu}{\sigma} > \dfrac{X - \mu}{\sigma} )[/tex]
[tex]P(X > 0.08) = P(z > \dfrac{0.08 - 0.07}{0.01} )[/tex]
[tex]P(X > 0.08) = P(z > \dfrac{0.01}{0.01} )[/tex]
[tex]P(X > 0.08) = P(z > 1 )[/tex]
[tex]P(X> 0.08) = 1- P(z < 1 )[/tex]
P(X > 0.08) = 1 - 0.8413
P(X > 0.08) = 0.1587
P(X > 0.08) [tex]\simeq[/tex] 16%
3. What percent of students who drink five beers have a BAC above 0.10 (the legal limit for driving in most states)?
[tex]P(X> 0.10) = P(\dfrac{0.10 - \mu}{\sigma} > \dfrac{X - \mu}{\sigma} )[/tex]
[tex]P(X > 0.10) = P(z > \dfrac{0.10 - 0.07}{0.01} )[/tex]
[tex]P(X > 0.10) = P(z > \dfrac{0.03}{0.01} )[/tex]
[tex]P(X > 0.10) = P(z > 3)[/tex]
[tex]P(X> 0.10) = 1- P(z < 3 )[/tex]
P(X > 0.10) = 1 - 0.9987
P(X > 0.08) = 0.0013
P(X > 0.08) [tex]\simeq[/tex] 0.15% which is the closet value to 0.0013