Question

The annual per capita consumption of bottled water was 30.3

gallons. Assume that the per capita consumption of bottled water is approximately normally distributed with a mean of

30.3 and a standard deviation of 10 gallons.

a. What is the probability that someone consumed more than

40 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. 99.59% of people consumed less than how many gallons of bottled​ water?

please solve A-D. Thankyou. Round to four decimal places

Answer 1

Solution :

Given that ,

mean = = 30.3

standard deviation = = 10

a.

P(x > 40) = 1 - P(x < 40)

= 1 - P[(x - ) / < (40 - 30.3) / 10)

= 1 - P(z < 0.97)

= 1 - 0.834

= 0.1660

Probability = 0.1660

b.

P(20 < x < 30) = P[(20 - 30.3)/ 10) < (x - ) /  < (30 - 30.3) / 10) ]

= P(-1.03 < z < -0.03)

= P(z < -0.03) - P(z < -1.03)

= 0.488 - 0.1515

= 0.3365

Probability = 0.3365

c.

P(x < 20) = P[(x - ) / < (20 - 30.3) / 10]

= P(z < -1.03)

= 0.1515

Probability = 0.1515

d.

Using standard normal table ,

P(Z < z) = 99.59%

P(Z < 2.644) = 0.9959

z = 2.644

Using z-score formula,

x = z * +

x = 2.644 * 10 + 30.3 = 56.74

56.74 gallons of bottled​ water