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*Please Provide step by step work* A poll Agency reports that 51% of teenagers a

ID: 3134949 • Letter: #

Question

*Please Provide step by step work*

A poll Agency reports that 51% of teenagers aged 12-17 played video games on their cell phone. A random sample of 151 teenagers is drawn. Use table A.2 if needed to round your answers to 4 decimal places.

1. The mean u p is_____

2. The standard deviation is _______

3. The probability that more than 53% of the teenagers in the sample play video games on their cell phone is_______

4. The the probability that the sample proportion who play video games of their cell phone is between 0.46 and 0.56 is ______

5. The probability that the sample proportion who play the video game on their cell phone is less than 0.57 is _____

6. It (would/ would not) be unusual if the sample proportion hwo play video games on their ell phones was less than 0.44, since the probability is ______

Explanation / Answer

1. The mean u p is__0.51___.

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2. The standard deviation is _______

Here,          
n =    151      
p =    0.51      

Hence,
          
s = standard deviation = sqrt(p(1-p)/n) =    0.040681284   [ANSWER]

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3. The probability that more than 53% of the teenagers in the sample play video games on their cell phone is_______

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.53      
u = mean = p =    0.51      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.040681284      
          
Thus,          
          
z = (x - u) / s =    0.491626564      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.491626564   ) =    0.31149168 [ANSWER]

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4. The the probability that the sample proportion who play video games of their cell phone is between 0.46 and 0.56 is ______

Here,          
n =    151      
p =    0.51      
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.46      
x2 = upper bound =    0.56      
u = mean = p =    0.51      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.040681284      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.229066411      
z2 = upper z score = (x2 - u) / s =    1.229066411      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.109523454      
P(z < z2) =    0.890476546      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.780953092   [ANSWER]

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5. The probability that the sample proportion who play the video game on their cell phone is less than 0.57 is _____

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.57      
u = mean = p =    0.51      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.040681284      
          
Thus,          
          
z = (x - u) / s =    1.474879693      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   1.474879693   ) =    0.929877555 [ANSWER]

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6. It (would/ would not) be unusual if the sample proportion hwo play video games on their ell phones was less than 0.44, since the probability is ______

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.44      
u = mean = p =    0.51      
          
s = standard deviation = sqrt(p(1-p)/n) =    0.040681284      
          
Thus,          
          
z = (x - u) / s =    -1.720692975      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.720692975   ) =    0.042653276

Hence,

It [[WOULD]] be unusual if the sample proportion hwo play video games on their ell phones was less than 0.44, since the probability is __0.042653276____ [which is less than 0.05].