Suppose we toss a fair coin 200 times Fill in the blanks Use

Suppose we toss a fair coin 200 times. Fill in the blanks: Use the Normal approximation to find the probability that the sample proportion is A) Between 0.4 and 0.6 is _____ (Give your answer to 4 decimal places). B) Between 0.45 and 0.55 is _____ (Give your answer to 4 decimal places).

Solution

Here, the mean proportion is

u = p = 0.5

and the standard deviation of the proportion is

sigma = sqrt(p(1-p)/n) = sqrt(0.5*(1-0.05)/200) = 0.048733972

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.4      
x2 = upper bound =    0.6      
u = mean =    0.5      
          
s = standard deviation =    0.048733972      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.051956693      
z2 = upper z score = (x2 - u) / s =    2.051956693      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.020086936      
P(z < z2) =    0.979913064      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.959826129   [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.45      
x2 = upper bound =    0.55      
u = mean =    0.5      
          
s = standard deviation =    0.048733972      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.025978346      
z2 = upper z score = (x2 - u) / s =    1.025978346      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.152450895      
P(z < z2) =    0.847549105      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.695098209   [ANSWER]