hello how do I solve 17 with work shown thank you The Instit

The Institute for College Access and Success reported that 67% of college students in a recent year graduated with student loan debt. A random sample of 85 graduates is drawn. Find the mean mu_p. Find the standard deviation sigma_p. Find the probability that less than 60% of the people in the sample were in debt. Find the probability that between 65% and 80% of the people in the sample were in debt. Find the probability that more than 75% of the people m the sample were in debt. Would it be unusual if less than 66% of the people m the sample were in debt?

Solution

a)

up = 0.67 [ANSWER]

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B)

sigmap = sqrt(p(1-p)/n) = sqrt(0.67*(1-0.67)/85) = 0.05100173 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.6      
u = mean =    0.67      
          
s = standard deviation =    0.05100173      
          
Thus,          
          
z = (x - u) / s =    -1.372502462      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.372502462   ) =    0.084953536 [ANSWER]

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d)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.65      
x2 = upper bound =    0.8      
u = mean =    0.67      
          
s = standard deviation =    0.05100173      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.392143561      
z2 = upper z score = (x2 - u) / s =    2.548933144      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.347476072      
P(z < z2) =    0.994597349      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.647121278   [ANSWER]

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e)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.75      
u = mean =    0.67      
          
s = standard deviation =    0.05100173      
          
Thus,          
          
z = (x - u) / s =    1.568574242      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.568574242   ) =    0.058373589 [ANSWER]

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f)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.65      
u = mean =    0.67      
          
s = standard deviation =    0.05100173      
          
Thus,          
          
z = (x - u) / s =    -0.392143561      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.392143561   ) =    0.347476072

NO. This is a large probability, it is not unusual. [ANSWER]