The breakdown voltage of a randomly chosen diode of a certai

The breakdown voltage of a randomly chosen diode of a certain type is known to be normally distributed with mean 40 V and standard deviation 1.5 V. What is the probability that the voltage of a single diode is between 3V and 42? P(39

Solution

A)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    39      
x2 = upper bound =    42      
u = mean =    40      
          
s = standard deviation =    1.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.666666667      
z2 = upper z score = (x2 - u) / s =    1.333333333      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.252492538      
P(z < z2) =    0.90878878      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.656296243   [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    1 - 0.15 = 0.85      
          
Then, using table or technology,          
          
z =    1.036433389      
          
As x = u + z * s,          
          
where          
          
u = mean =    40      
z = the critical z score =    1.036433389      
s = standard deviation =    1.5      
          
Then          
          
x = critical value =    41.55465008   [ANSWER]

This is the 85th percentile.

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c)

i)

It has the same mean

u(X) = 40

and variance of

sigma^2(X) = sigma^2/n = 1.5^2/9 = 0.25 [ANSWER]

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ii)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    39.5      
x2 = upper bound =    41      
u = mean =    40      
          
s = standard deviation =    0.25      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2      
z2 = upper z score = (x2 - u) / s =    4      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.999968329      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.977218197   [ANSWER]