Question

A ski gondola carries skiers to the top of a mountain. Assume that weights of skiers are normally distributed with a mean of

180

lb and a standard deviation of

37

lb. The gondola has a stated capacity of

25

​passengers, and the gondola is rated for a load limit of

3500

lb. Complete parts​ (a) through​ (d) below.a. Given that the gondola is rated for a load limit of

3500

​lb, what is the maximum mean weight of the passengers if the gondola is filled to the stated capacity of

25

​passengers?The maximum mean weight is

nothing

lb.

​(Type an integer or a decimal. Do not​ round.)

b. If the gondola is filled with

25

randomly selected​ skiers, what is the probability that their mean weight exceeds the value from part​ (a)?The probability is

nothing.

​(Round to four decimal places as​ needed.)

c. If the weight assumptions were revised so that the new capacity became

20

passengers and the gondola is filled with

20

randomly selected​ skiers, what is the probability that their mean weight exceeds

175

​lb, which is the maximum mean weight that does not cause the total load to exceed

3500

​lb?The probability is

nothing.

​(Round to four decimal places as​ needed.)

d. Is the new capacity of

20

passengers​ safe?Since the probability of overloading is

▼

over 50 % commaover 50%,

under 5 % commaunder 5%,

the new capacity

▼

appearsappears

does not appeardoes not appear

to be safe enough.

Answer 1

solution:

Given that mean = 180 lb , standard deviation = 37 lb

a) maximum mean weight of passengers = load limit /number of passengers

The maximum mean weight of passengers = 3500/25 = 140 lb

b)First, find the z-score: z = (value - mean) / (stdev/sqrt(n))
=> P(X > 140) = P(Z > (140-180)/(37/sqrt(25)))
= P(Z > -5.4054)
= 1

c. With only 20 passengers, the new maximum mean weight of passengers = 3500 ÷ 20 = 175lb

=> P(X > 175) = P(Z > (175-180)/(37/sqrt(20)))
= P(Z > -0.6043)
= 0.7257

d)Since the probability of overloading is under 5% the new capacity over 50% to be safe enough.