Question

 

In 2012, the per capita consumption of soft drinks in the United States was reported to be 44 gallons. Assume that the per capita consumption of soft drinks in the USA is approximately normally distributed with a mean of 44 gallons and a standard deviation of 14 gallons.

What is the probability that someone in the United States consumed more than 70 gallons of soft drinks in 2012?

1.             0.3752
2.             0.1395
3.             0.2588
4.             0.0316

Refer back to the soft drink consumption data provided earlier. What is the probability that someone in the USA consumed between 20 and 30 gallons of soft drinks in 2012?

1.             0.1265
2.             0.3752
3.             0.0316
4.             0.1154

Refer back to the soft drink consumption data provided earlier. What is the probability that someone in the USA consumed less than 15 gallons of soft drinks in 2012?

1.             0.1265
2.             0.0192
3.             0.0543
4.             0.1395

Answer 1

Solution :

Given that,

mean = = 44

standard deviation = =14

(A_P(x >70 ) = 1 - P(x< 70)

= 1 - P[(x -) / < (70-44) /14 ]

= 1 - P(z <1.8571 )

Using z table

= 1 -  0.9684

probability=0.0316

(B)

P(20< x <30 ) = P[(20-44) / 14< (x - ) / < (30-44) / 14)]

= P(-1.7143 < Z < -1)

= P(Z <-1 ) - P(Z <-1.7143 )

Using z table   

= 0.1587-0.0432

probability=0.1154

(D)

P(X<15 ) = P[(X- ) / < (15-44) /14 ]

= P(z <-2.0714 )

Using z table

=0.0192

probability=0.0192