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 6.00 hours per week playing organized sports. The population\'s standard deviation is 3.40 hours per week. Based on a sample of 49 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 5.4 and 6.6 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 5.6 and 6.4 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)

= 3.4/sqrt(49)

= 0.485714286 [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 =    5.4      
x2 = upper bound =    6.6      
u = mean =    6      
n = sample size =    49      
s = standard deviation =    3.4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.235294118      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.235294118      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.108360531      
P(z < z2) =    0.891639469      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.783278938   [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 =    5.6      
x2 = upper bound =    6.4      
u = mean =    6      
n = sample size =    49      
s = standard deviation =    3.4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.823529412      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.823529412      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.205103499      
P(z < z2) =    0.794896501      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.589793001   [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 =    8.4      
u = mean =    6      
n = sample size =    49      
s = standard deviation =    3.4      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    4.941176471      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   4.941176471   ) =    0.000000388263 [VERY UNLIKELY!]