In the Department of Education at UR University student reco

In the Department of Education at UR University, student records suggest that the population of students spends an average of 6.50 hours per week playing organized sports. The population\'s standard deviation is 3.50 hours per week. Based on a sample of 64 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates.

What is the chance HLI will find a sample mean between 5.6 and 7.4 hours? (Round z and standard error values to 2 decimal places and final answer to 4 decimal places.)

Calculate the probability that the sample mean will be between 6.1 and 6.9 hours. (Round z and standard error values to 2 decimal places and final answer to 4 decimal places.)

Solution

a)

standard error = s/sqrt(n)

= 3.50/sqrt(64)

= 0.4375 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    5.6      
x2 = upper bound =    7.4      
u = mean =    6.5      
n = sample size =    64      
s = standard deviation =    3.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.057142857      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.057142857      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.019836243      
P(z < z2) =    0.980163757      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.960327514   [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    6.1      
x2 = upper bound =    6.9      
u = mean =    6.5      
n = sample size =    64      
s = standard deviation =    3.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.914285714      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.914285714      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.180283368      
P(z < z2) =    0.819716632      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.639433265   [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    8.7      
u = mean =    6.5      
n = sample size =    64      
s = standard deviation =    3.5      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    5.028571429      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   5.028571429   ) =    2.47074*10^-7 [VERY UNLIKELY]