In a random sample of 150 community college students, the mean number of hours s

Question

In a random sample of 150 community college students, the mean number of hours spent studying per week is 11.7 hours and the standard deviation is 4 hours.

a) Find the standard score (z-score) for students who study the following hours in a particular week. Round Z to the nearest hundredth and interpret the meaning of each answer as it pertains to this problem. i) 22 hours ii) 6 hours

b)Assuming the distribution of the number of hours community college students study per week is normally distributed, approximately how many of the students in the sample study between 7.7 and 15.7 hours per week?

c) Assuming the distribution of the number of hours community college students study per week is normally distributed, approximately how many of the students in the sample study less than 3.7 hours per week?

d) Without assuming anything about the distribution of the number of hours community college students study per week, at least what percentage (approximately) of the students study between 5.3 and 18.1 hours per week?

e) Without assuming anything about the distribution of the number of hours community college students study per week, at most what percentage of the students study less than 3.7 hours or more than 19.7 hours per week?

Explanation / Answer

a)

i)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    22      
u = mean =    11.7      
          
s = standard deviation =    4      
          
Thus,          
          
z = (x - u) / s =    2.575 = 2.58 [ANSWER]

This means that this student is 2.58 standard deviations above the mean when it comes to hours spent in studying.

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ii)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    6      
u = mean =    11.7      
          
s = standard deviation =    4      
          
Thus,          
          
z = (x - u) / s =    -1.425 = -1.43 [ANSWER]

This means that this student is 1.43 standard deviations below the mean when it comes to hours spent in studying.

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    7.7      
x2 = upper bound =    15.7      
u = mean =    11.7      
          
s = standard deviation =    4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.682689492      

Hence, around 0.68268*150 = 102.402 or 102 students [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 =    3.7      
u = mean =    11.7      
          
s = standard deviation =    4      
          
Thus,          
          
z = (x - u) / s =    -2      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2   ) =    0.022750132

Hence, around 0.02275*150 = 3.41 or 3 students [ANSWER]

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