At the end of the spring semester, the Dean of Students sent a survey to the ent
ID: 3130465 • Letter: A
Question
At the end of the spring semester, the Dean of Students sent a survey to the entire freshman class. One question asked the students how much weight they had gained or lost since the beginning of the school year. The average was a gain of µ = 9 lbs. and = 6. The distribution of scores was approximately normal. A sample of n = 9 students is selected and the average weight change is computed for the sample. Answer the following questions: What is the probability that the sample mean will be less than M = 10 lbs. 2 pts.
Of all possible samples, what proportion shows an average weight gain of less than 5 lbs.? 2 pts.
What is the probability that the sample mean will be a gain between 4 and 12 lbs.? 4 pts.
What is the probability of selecting a student with a weight gain (score) over 15 lbs.? 2 pts.
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 10
u = mean = 9
n = sample size = 9
s = standard deviation = 6
Thus,
z = (x - u) * sqrt(n) / s = 0.5
Thus, using a table/technology, the left tailed area of this is
P(z < 0.5 ) = 0.691462461 [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 = 5
u = mean = 9
n = sample size = 9
s = standard deviation = 6
Thus,
z = (x - u) * sqrt(n) / s = -2
Thus, using a table/technology, the left tailed area of this is
P(z < -2 ) = 0.022750132 [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 = 4
x2 = upper bound = 12
u = mean = 9
n = sample size = 9
s = standard deviation = 6
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -2.5
z2 = upper z score = (x2 - u) * sqrt(n) / s = 1.5
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.006209665
P(z < z2) = 0.933192799
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.926983133 [ANSWER]
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d)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 15
u = mean = 9
s = standard deviation = 6
Thus,
z = (x - u) / s = 1
Thus, using a table/technology, the right tailed area of this is
P(z > 1 ) = 0.158655254 [ANSWER]
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