Suppose we are interested in the amount of time college stud

Suppose we are interested in the amount of time college students spend on social networking sites: Facebook, MySpace, and Twitter. We know the amount of time each student spends on social networking sites follows a Normal distribution with mean 7.6 hours and standard deviation 2.2. In this study we examine a sample of 20 college students. What is the probability that a student will spend more than 9 hours on social networking sites per week? What is the probability that the average amount of time spent on social networking sites is greater than 9 hours per week? Suppose the sample mean of hours spent on social networking sites per week was 11.5 hours. Is this above or below the 90th percentile of mean time spent on social networking sites? The number of bicycle versus automobile accidents per week at the intersection of Pardall Road and Embarcadero Del Norte varies with mean 3.2 and standard deviation 1.4. (This distribution is discrete and so it is not normal.) Let the sample mean, X, be the mean number of accidents per week at the intersection during a year (52 weeks). What is the distribution of the sample mean? What is the probability that the mean number of accidents is less than 2? What is the probability that there are on average more than 5 accidents per week?

Solution

4.

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    9      
u = mean =    7.6      
          
s = standard deviation =    2.2      
          
Thus,          
          
z = (x - u) / s =    0.636363636      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.636363636   ) =    0.262269718 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    9      
u = mean =    7.6      
n = sample size =    20      
s = standard deviation =    2.2      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.845904699      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.845904699   ) =    0.002214272 [ANSWER]

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C)

For the 90th percentile:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    7.6      
z = the critical z score =    1.281551566      
s = standard deviation =    2.2      
n = sample size =    20      
Then          
          
x = 90th percentile =    8.230440012      

Thus, 11.5 hours is ABOVE the 90th oercentile of means. [ANSWER]
          

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