The president of Doerman Distributors, Inc., believes that 31% of the firm’s ord

Question

The president of Doerman Distributors, Inc., believes that 31% of the firm’s orders come from first-time customers. A random sample of 101 orders will be used to estimate the proportion of first-time customers. Assume that the president is correct and

1. Compute the standard error of the sample proportion (provide your answer using exactly three decimal places).

2. What is the probability that the sample proportion will be between 0.21 and 0.41? (provide your answer using exactly four decimal places)

3. What is the probability that the sample proportion will be between 0.26 and 0.36? (provide your answer using exactly four decimal places)

4. What is the probability that the sample proportion will be greater than 0.38? (provide your answer using exactly four decimal places)

Explanation / Answer

1.

Here, the proportion of p has a mean and standard deviation of

u(p) = 0.31

s(p) = sqrt(p(1-p)/n) = sqrt(0.31*(1-0.31)/101) = 0.046019798 [ANSWER, STANDARD ERROR]

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2)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.21      
x2 = upper bound =    0.41      
u = mean =    0.31      
          
s = standard deviation =    0.046019798      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.172977813      
z2 = upper z score = (x2 - u) / s =    2.172977813      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.014890994      
P(z < z2) =    0.985109006      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.970218013   [ANSWER]  

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3)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.26      
x2 = upper bound =    0.36      
u = mean =    0.31      
          
s = standard deviation =    0.046019798      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.086488906      
z2 = upper z score = (x2 - u) / s =    1.086488906      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.138631373      
P(z < z2) =    0.861368627      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.722737254   [ANSWER]

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4)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.38      
u = mean =    0.31      
          
s = standard deviation =    0.046019798      
          
Thus,          
          
z = (x - u) / s =    1.521084469      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.521084469   ) =    0.064119321 [ANSWER]
  

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