Question

The annual per capita consumption of bottled water was 33.1 gallons. Assume that the per capita consumption of bottled water is approximately normally distributed with a mean of 33.1 and a standard deviation of 11 gallons.

a. What is the probability that someone consumed more than 38 gallons of bottled​ water?

b. What is the probability that someone consumed between 30 and 40 gallons of bottled​ water?

c. What is the probability that someone consumed less than 30 gallons of bottled​ water?

d. 90% of people consumed less than how many gallons of bottled​ water?

Answer 1

Solution :

Given that ,

mean = = 33.1

standard deviation = = 11

a)

P(x > 38) = 1 - P(x < 38)

= 1 - P((x - ) / < (38 - 33.1) / 11)

= 1 - P(z < 0.45)

= 1 - 0.6736 Using standard normal table.

= 0.3264

Probability = 0.3264

b)

P(30 < x < 40) = P((30 - 33.1)/ 11) < (x - ) /  < (40 - 33.1) / 11) )

= P(-0.28 < z < 0.63)

= P(z < 0.63) - P(z < -0.28)

= 0.7357 - 0.3897 Using standard normal table,  

Probability = 0.3460

c)

P(x < 30) = P((x - ) / < (30 - 33.1 ) / 11)

= P(z < -0.28)

= 0.3897 Using standard normal table,

Probability = 0.3897

d)

The z - distribution of the90 % is,

P( Z < z ) = 90 %

P( Z < z ) = 0.90

P( Z < 1.282 ) = 0.90

z = 1.282

Using z - score formula,

X = z * +

= 1.282 * 11 + 33.1  

= 47.202