Show all Work c A tire manufacturer designed a new tread pat

Show all Work

(c). A tire manufacturer designed a new tread pattern for its all-weather tires. Repeated tests were conducted on cars of approximately the same weight travelling at 60 miles per hour. The test showed that the new tread pattern enables the cars to stop completely in an average distance of 125 feet with a standard deviation of 6.5 feet and that the stopping distances are approximately normally distributed.

(i). What is the 70th percentile of the distribution of stopping distances? (5 points)

(ii).What is the probability that at least 2 cars out of 5 randomly selected cars in the study will stop in a distance that is greater than the distance calculated in part (i)? (5 points)

(iii). What is the probability that a randomly selected sample of 5 cars in the study will have a mean stopping distance of at least 130 feet? (5 points)

Solution

I)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.7      
          
Then, using table or technology,          
          
z =    0.524400513      
          
As x = u + z * s,          
          
where          
          
u = mean =    125      
z = the critical z score =    0.524400513      
s = standard deviation =    6.5      
          
Then          
          
x = critical value =    128.4086033   [ANSWER]  

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ii)

The probability that it exceeds that distance is 1 - 0.70 = 0.30.

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    5      
p = the probability of a success =    0.3      
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.52822
          
Thus, the probability of at least   2   successes is  
          
P(at least   2   ) =    0.47178 [ANSWER]

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iii)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    130      
u = mean =    125      
n = sample size =    5      
s = standard deviation =    6.5      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.72005229      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.72005229   ) =    0.042711468 [ANSWER]