Mathematics, 27.07.2019 06:30 mohammedel04

Aski 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.

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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 from: silveryflight

Maximum Mean weight of the passenger filled to the stated capacity of 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 randomly selected skiers exceeds $150\;lb$ is about $88.5\;%$ %

C) With only 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 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 skiers is safer than skiers, but we cannot consider it safe enough, since the maximum capacity should be of skiers.

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