According to a recent survey, about 33% of Americans polled said that they would

Question

According to a recent survey, about 33% of Americans polled said that they would likely purchase reusable cloth bags for groceries in order to reduce plastic waste. Suppose 45 shoppers are interviewed a local supermarket:

a. Describe the sampling distribution for the sample proportion by checking the Normality of the model and determining its mean and standard deviation.

b. What is the probability that no more than 28% of shoppers say that they are likely to purchase reusable cloth bags for groceries?

c. What is the probability that between 28% and 38% of shoppers say that they are likely to purchase reusable cloth bags for groceries?

Explanation / Answer

a)

Here,          
n =    45      
p =    0.33      

As np > 10, n(1-p) > 10, then we can approximate the distribution as normal.

Hence,

u = mean = p =    0.33      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.070095173   [ANSWER]
******************************

b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.28      
u = mean = p =    0.33      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.070095173      
          
Thus,          
          
z = (x - u) / s =    -0.713315876      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.713315876   ) =    0.237825158 [ANSWER]

**************************

c)

Here,          
n =    45      
p =    0.33      
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.28      
x2 = upper bound =    0.38      
u = mean = p =    0.33      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.070095173      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.713315876      
z2 = upper z score = (x2 - u) / s =    0.713315876      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.237825158      
P(z < z2) =    0.762174842      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.524349684   [ANSWER]     

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.