Question

Assume that the length of time for a horseback ride on the trail at Triple R Ranch is normally distributed with a mean of 3.2 hours and a standard deviation of 0.4 hours.

a) What percent of horseback rides last at least 3.2 hours?

b) What percent of horseback rides last less than 2.8 hours?

c) What percent of horseback rides last at least 3.7 hours?

d) What percent of horseback rides are between 2.8 hours and 4.0 hours?

e) In a random sample of 500 horseback rides at Triple R Ranch, how many are at least 3.7 hours?

Answer 1

Solution :

Given that ,

mean = = 3.2

standard deviation = = 0.4

a)P(x 3.2 ) = 1 - P(x   3.2)

= 1 - P[(x - ) / (3.2 -3.2) / 0.4]

= 1 -  P(z 0 )  

= 1 -0.5 = 0.50

Probability = 0.5000 = 50%

b)P(x < 2.8 ) = P[(x - ) / < (2.8 -3.2) /0.4 ]

= P(z < -1)

= 0.1587

probability = 0.1587 = 15.87%

c)P(x 3.7 ) = 1 - P(x   3.7)

= 1 - P[(x - ) / (3.7 -3.2) / 0.4]

= 1 -  P(z 1.25 )  

= 1 - 0.8944 = 0.1056

Probability = 0.1056 = 10.56%

d)

P(2.8 < x <4.0 ) = P[(2.8 -3.2)/0.4 ) < (x - ) /  < (4 -3.2) / 0.4) ]

= P(-1 < z < 2)

= P(z < 2) - P(z < -1)

Using standard normal table

= 0.9772 - 0.1587 = 0.8185

Probability =0.8185 = 81.85%

e)

P(x 3.7 ) = 1 - P(x   3.7)

= 1 - P[(x - ) / (3.7 -3.2) / 0.4]

= 1 -  P(z 1.25 )  

= 1 - 0.8944 = 0.1056

Probability = 0.1056*500 = 52.8 hours.