Question

A ski gondola carries skiers to the top of a mountain. assume that weights of skiers are normally distributed with a mean of 199 lb and a standard deviation of 41 lb. the gondola has a stated capacity of 25 ​passengers, and the gondola is rated for a load limit of 3750 lb. complete parts​ (a) through​ (d) below.
a. given that the gondola is rated for a load limit of 3750 ​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 150 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 1. ​(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 187.5 ​lb, which is the maximum mean weight that does not cause the total load to exceed 3750 ​lb? the probability is . 8944. ​(round to four decimal places as​ needed.)
d. is the new capacity of 20 passengers​ safe? since the probability of overloading is over 50 % comma the new capacity appears to be safe enough.

Answer 1

A) maximum mean weight of passengers = load limit ÷ number of passengers

maximum mean weight of passengers = 3750 ÷ 25 = 150lb

B)  First, find the z-score:
z = (value - mean) / stdev
   = (150 - 199) / 41
   = -1.20

We need to find P(z > -1.20) = 1 - P(z < -1.20)

Now, look at a standard normal table to find P(z < -1.20) = 0.11507, therefore:
P(z > -1.20) = 1 - 0.11507 = 0.8849

Hence, the probability that the mean weight of 25 randomly selected skiers exceeds 150lb is about 88.5% 

C) With only 20 passengers, the new maximum mean weight of passengers = 3750 ÷ 20 = 187.5lb

Let's repeat the steps of point B)

z = (187.5 - 199) / 41
   = -0.29

P(z > -0.29) = 1 - P(z < -0.29) = 1 - 0.3859 = 0.6141

Hence, the probability that the mean weight of 20 randomly selected skiers exceeds 187.5lb is about 61.4%

D) The mean weight of skiers is 199lb, therefore:
number of passengers = load limit ÷ mean weight of passengers
                                     = 3750 ÷ 199
                                     = 18.8
The new capacity of 20 skiers is safer than 25 skiers, but we cannot consider it safe enough, since the maximum capacity should be of 18 skiers.

Answer 2

Maximum Mean weight of the passenger filled to the stated capacity of $25$ ​passengers is $150lb$

A) Maximum mean weight of passengers = $load \; limit \div number\; of\; passengers$

Maximum mean weight of passengers = $3750 \div 25$

Maximum mean weight of passengers$=150lb$

B)  First, find the z-score:

$z=\dfrac{value-mean}{standard\;deviation}$

$z=\dfrac{150-199}{41}$

$z=-1.20$

We need to find $P(z > -1.20)= 1 - P(z < -1.20)$

Now, look at a standard normal table to find $P(z < -1.20) = 0.1150$ , therefore:

$P(z > -1.20) = 1 - 0.11507 = 0.8849$

Hence, the probability that the mean weight of $25$ randomly selected skiers exceeds $150\;lb$ is about $88.5\;%$%

C) With only $20$ passengers, the new maximum mean weight of passengers =$3750 \div 20=187.5\;lb$

Repeat the steps of point B)

$z=\dfrac{187.5-199}{41}$

$z=-0.29$

$P(z > -0.29) = 1 - P(z < -0.29) \\P(z > -0.29) = 1 - 0.3859 \\P(z > -0.29) = 0.6141$

Hence, the probability that the mean weight of $20$ randomly selected skiers exceeds $187.5\;lb$ is about $61.4\;%$%

D) The mean weight of skiers is $199\;lb$ , therefore:

number of passengers = $load \;limit \div mean\; weight \;of \; passengers$l                            $number\; of\; passengers=3750\div199$

$number\; of\; passengers=18.8$

The new capacity of $20$ skiers is safer than $25$ skiers, but we cannot consider it safe enough, since the maximum capacity should be of $18$ skiers.

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