Assume that the population proportion of delayed flights during bad weather cond

Question

Assume that the population proportion of delayed flights during bad weather conditions is p=0.25. a. What is the sampling distribution of p-bar, the proportion of late flights in a sample of 1,000 flights taken during bad weather conditions? b. What is the probability that the sample proportion will be within +/- .03 of the population proportion if a sample size of 1,000 is selected? c. What is the probability that the sample proportion will be within +/- .03 of the population proportion if a sample size of 500 is selected?

Explanation / Answer

a)

Here,  
n =    1000
p =    0.25

u = mean = p =    0.25
  
s = standard deviation = sqrt(p(1-p)/n) =    0.013693064

Hence, it is approximately normal with mean 0.25 and standard deviation of 0.013693064. [ANSWER]

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b)

That means between 0.22 and 0.28.

Here,          
n =    1000      
p =    0.25      
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.22      
x2 = upper bound =    0.28      
u = mean = p =    0.25      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.013693064      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.19089023      
z2 = upper z score = (x2 - u) / s =    2.19089023      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.014229868      
P(z < z2) =    0.985770132      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.971540263   [ANSWER]

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c)

Here,          
n =    500      
p =    0.25      
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.22      
x2 = upper bound =    0.28      
u = mean = p =    0.25      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.019364917      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.549193338      
z2 = upper z score = (x2 - u) / s =    1.549193338      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.060667625      
P(z < z2) =    0.939332375      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.87866475   [ANSWER]

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