A bench test for determining maximum speed of a Moped has a

A bench test for determining maximum speed of a Moped has a normal distribution with mean value 46.8 km/h and st. deviation 1.75 km/h.

a) If a single Moped is selected, what is the probability that maximum speed is at least 48 km/h?

b) For a sample of 50 randomly selected Mopeds, what is the probability that the sample mean X is between 45 km/h and 50 km/h?

Solution

a)
We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    48      
u = mean =    46.8      
          
s = standard deviation =    1.75      
          
Thus,          
          
z = (x - u) / s =    0.685714286      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.685714286   ) =    0.246446648 [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 =    45      
x2 = upper bound =    50      
u = mean =    46.8      
n = sample size =    50      
s = standard deviation =    1.75      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -7.273098321      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    12.92995257      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    1.75667E-13      
P(z < z2) =    1      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    1   [ANSWER]