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

196196

lb and a standard deviation of

3939

lb. The gondola has a stated capacity of

2525

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

37503750

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

37503750

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

2525

​passengers?The maximum mean weight is

150150

lb.

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

b. If the gondola is filled with

2525

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

11 .

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

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

2020

passengers and the gondola is filled with

2020

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

187.5187.5

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

37503750

​lb?The probability is

nothing .

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

d. Is the new capacity of

2020

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

Mean = 196

standard deviation = 39

The gondola has a stated capacity =25

the gondola is rated for a load limit = 3750

a)

The maximum mean weight is = 3750/25 = 150

b)

Probability when weight greater than 150

µ =    196                                  
σ =    39                                  
n=   20                                  
                                      
X =   150                                  
                                      
Z =   (X - µ )/(σ/√n) = (   150   -   196   ) / (    39   / √   20   ) =   -5.275
                                      
P(X ≥   150   ) = P(Z ≥   -5.27   ) =   P ( Z <   5.275   ) =    1.0000      
excel formula for probability from z score is =NORMSDIST(Z)                                      

c)

Probability when weight greater than 187.5

µ =    196                                  
σ =    39                                  
n=   20                                  
                                      
X =   187.5                                  
                                      
Z =   (X - µ )/(σ/√n) = (   187.5   -   196   ) / (    39   / √   20   ) =   -0.975
                                      
P(X ≥   187.5   ) = P(Z ≥   -0.97   ) =   P ( Z <   0.975   ) =    0.8351      
excel formula for probability from z score is =NORMSDIST(Z)                                      

d)

Over 50%

does not appears to be safe

Please revert back in case of any doubt.

Please upvote. Thanks in advance.