with the method that we should useSolution1 a standard error

with the method that we should use

Solution

1.

a)

standard error (se) = s / sqrt(n) = 100/sqrt(60) = 12.90994449 [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 =    477      
x2 = upper bound =    527      
u = mean =    502      
n = sample size =    60      
s = standard deviation =    100      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.936491673      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.936491673      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.026403756      
P(z < z2) =    0.973596244      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.947192489   [answer]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    492      
x2 = upper bound =    512      
u = mean =    502      
n = sample size =    60      
s = standard deviation =    100      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.774596669      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.774596669      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.219289013      
P(z < z2) =    0.780710987      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.561421974   [answer]

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d)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    550      
u = mean =    502      
n = sample size =    60      
s = standard deviation =    100      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    3.718064012      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3.718064012   ) =    0.000100378 [answer]

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with the method that we should useSolution1. a) standard error (se) = s / sqrt(n) = 100/sqrt(60) = 12.90994449 [answer] ************ b) We first get the z score
with the method that we should useSolution1. a) standard error (se) = s / sqrt(n) = 100/sqrt(60) = 12.90994449 [answer] ************ b) We first get the z score

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