Chapter 6 5.) DMV reports that the average age of a vehicle in Santa Clara is 9

Question

Chapter 6

5.) DMV reports that the average age of a vehicle in Santa Clara is 9 years old (108 months). Assume that the distribution of vehicle ages is normally distributed with a standard deviation of 18 months. What percent of vehicles in Santa Clara are between 10 (120 months) and 15 years old (180 months). (Express your answer as a decimal rounded to four decimal places)

Part II The DMV reports that the average age of a vehicle in Santa Clara is 9 years old (108 months) with a standard deviation of 18 months. For a random sample of 20 vehicles in Santa Clara what is the probability that the average age of the vehicles in the sample is between 10 (120 months) and 15 years old (180 months). Give your answer as a decimal rounded to the fourth decimal place.

Explanation / Answer

5.

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    120      
x2 = upper bound =    180      
u = mean =    108      
          
s = standard deviation =    18      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.666666667      
z2 = upper z score = (x2 - u) / s =    4      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.747507462      
P(z < z2) =    0.999968329      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.252460866 = 25.246% [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    120      
x2 = upper bound =    180      
u = mean =    108      
n = sample size =    20      
s = standard deviation =    18      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    2.98142397      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    17.88854382      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.998565444      
P(z < z2) =    1      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.001434556   [ANSWER]  
      

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