The times taken to complete an introduction to business statistics exam have a n

Question

The times taken to complete an introduction to business statistics exam have a normal distribution with a mean of 65 minutes and standard deviation of 7 minutes. There are 150 students who took the exam and students are allowed a total of 75 minutes to take the exam.

a) What is the chance that Mike finished his exam in 63 to 72 minutes?

b) What is the expected number of students who finished in less than 75 minutes?

c) As some students were not able to finish the exam in time, the instructor allowed 6 more minutes. Given he already spent 75 minutes on the exam, what is the chance that Chris finished his exam in extended time, that is between 75 and 81 minutes

Explanation / Answer

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    63      
x2 = upper bound =    72      
u = mean =    65      
          
s = standard deviation =    7      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.285714286      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.387548481      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.453796265   [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    75      
u = mean =    65      
          
s = standard deviation =    7      
          
Thus,          
          
z = (x - u) / s =    1.428571429      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   1.428571429   ) =    0.923436274

Hence, we expect 0.92343*150 = 138.5145 students [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    75      
x2 = upper bound =    81      
u = mean =    65      
          
s = standard deviation =    7      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    1.428571429      
z2 = upper z score = (x2 - u) / s =    2.285714286      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.923436274      
P(z < z2) =    0.988864511      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.065428236      

Hence,

P(75<x<81|x>75) = 0.065428236/(1-0.923436274) = 0.854559194 [ANSWER]

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