Loony Park has a new adult ride that holds 25 people The max

Loony Park has a new adult ride that holds 25 people. The maximum weight that the ride can carry is 4,255.001b. The weight of adults has a mean of 159.831b and a standard deviation of 18.91lb. Calculate the probability that a random group of 25 people will overload the maximum weight that the ride can carry. This is equivalent to calculating the probability that the mean weight of 25 adults would exceed 170.201b (being 4,255.001b divided by 25). You may find this standard normal table useful. Give your answer as a decimal to 4 decimal places. probability =

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    170.2      
u = mean =    159.83      
n = sample size =    25      
s = standard deviation =    18.91      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.741935484      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.741935484   ) =    0.003053917 [answer]

 Loony Park has a new adult ride that holds 25 people. The maximum weight that the ride can carry is 4,255.001b. The weight of adults has a mean of 159.831b and

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