20% of fine dinning restaurants have policies of restricting the use of cell pho

Question

20% of fine dinning restaurants have policies of restricting the use of cell phones. If you select a random sample of 100 fine dinning restaurants a.) What is the probability that the sample has between 15% and 25% that have established policies restricting cell phone use? b.) The probability is 90% that the sample percentage will be contained within what symmetrical limits of the population percentage? c.) Supposed that a year later, a random sample of 100 restaurants were selected and it was found out that 31% had policies restricting the use of cell phones. Do you think that the population percentage has changed?

Explanation / Answer

Here, the distribution of the proportion is

u = mean = 0.20
standard deviation = sqrt(p(1-p)/n) = sqrt(0.20*(1-0.20)/100) = 0.04

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.15      
x2 = upper bound =    0.25      
u = mean =    0.2      
          
s = standard deviation =    0.04      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.25      
z2 = upper z score = (x2 - u) / s =    1.25      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.105649774      
P(z < z2) =    0.894350226      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.788700453   [ANSWER]

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b)

The middle 90% is the interval between the 5th and 95th percentile.

5th percentile:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.05      
          
Then, using table or technology,          
          
z =    -1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    0.2      
z = the critical z score =    -1.644853627      
s = standard deviation =    0.04      
          
Then          
          
x = critical value =    0.134205855      

95th percentile:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    0.2      
z = the critical z score =    1.644853627      
s = standard deviation =    0.04      
          
Then          
          
x = critical value =    0.265794145

Thus, the interval is between 0.134205855 and 0.265794145. [ANSWER]

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c)

Yes, because 0.31 is outside the interval in part B.  

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