Suppose that 37% of children say that purple is their favorite color and suppose

Question

Suppose that 37% of children say that purple is their favorite color and suppose that 30 children are randomly selected. Use this information to answer the questions.

1 If we are using the normal approximation for this problem, what is the mean of the distribution?

2 If we are using the normal approximation for this problem, what is the standard deviation of the distribution (the standard error)? Round your answer to four decimal places.

3 What is the probability that at least 40% of the 30 children say that purple is their favorite color? Use the normal approximation. Round your answer to three decimal places.

4 What is the probability that no more than 30% of the 30 children say that their favorite color is purple?  Use the normal approximation.  Round your answer to three decimal places.

5 What is the probability that between 25% and 50% of the 30 children say that purple is their favorite color?  Use the normal approximation.  Round your answer to three decimal places.

Explanation / Answer

1.

As n = 30, p = 0.37,

u = mean = np =    11.1 [ANSWER]

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2.
  
s = standard deviation = sqrt(np(1-p)) =    2.644428105 [ANSWER]

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3.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.4      
u = mean = p =    0.37      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.088147603      
          
Thus,          
          
z = (x - u) / s =    0.340338237      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.340338237   ) =    0.366800912 [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 =    0.3      
u = mean = p =    0.37      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.088147603      
          
Thus,          
          
z = (x - u) / s =    -0.794122554      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.794122554   ) =    0.213562044 [ANSWER]

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5.

Here,          
n =    30      
p =    0.37      
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.25      
x2 = upper bound =    0.5      
u = mean = p =    0.37      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.088147603      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.36135295      
z2 = upper z score = (x2 - u) / s =    1.474799029      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.086701089      
P(z < z2) =    0.929866709      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.84316562   [ANSWER]  

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