Question #147933
he annual per capita consumption of bottled water was 31.9 gallons. Assume that the per capita consumption of bottled water is approximately normally distributed with a mean of 31.9 and a standard deviation of 10 gallons.
a. What is the probability that someone consumed more than 37 gallons of bottled​ water?
b. What is the probability that someone consumed between 20 and 30 gallons of bottled​ water?
c. What is the probability that someone consumed less than 20 gallons of bottled​ water?
d. 99​% of people consumed less than how many gallons of bottled​ water?
1
Expert's answer
2020-12-01T20:45:34-0500

(a)  P(x>37)=1P(x<37)=1P(xμσ<3731.910)=1P(z<0.51)=0.3050(a) \; P(x > 37) = 1 – P(x < 37) \\ = 1 – P(\frac{x – μ}{σ} < \frac{37 – 31.9}{10}) \\ = 1 – P(z < 0.51) \\ = 0.3050

(b)  P(20<x<30)=P(2031.910<xμσ<3031.910)=P(1.19<z<0.19)=P(z<0.19)P(z<1.19)=0.42460.1170=0.3076(b)\; P( 20 < x < 30) = P(\frac{20 – 31.9}{10} < \frac{x – μ}{σ} < \frac{30 – 31.9}{10}) \\ = P(-1.19 < z < -0.19) \\ = P(z < -0.19) – P(z < -1.19) \\ = 0.4246 – 0.1170 \\ = 0.3076

(c)  P(x<20)=P(xμσ<2031.910)=P(z<1.19)=0.1170(d)  P(z<2.33)=0.99z=2.33x=z×σ+μx=2.33×10+31.9=55.2(c) \; P(x < 20) = P(\frac{x – μ}{σ} < \frac{20 – 31.9}{10}) \\ = P(z < -1.19) \\ = 0.1170 \\ (d) \; P(z < 2.33) = 0.99\\ z = 2.33 \\ x = z \times σ + μ \\ x = 2.33 \times 10 + 31.9 = 55.2

Answer: 55.2 gallons


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