Homework SourseScores on a recent national statistics exam were normally di
Scores on a recent national statistics exam were normally distributed with a mean of 80 and standard deviation of 6. 1. What is the probability that a randomly selected exam will have a score of at least 71, 2. what percentage of exam will have scores between 89 and 92, and if the top 2.5 % of the teast scores receive merit awards, what is the lowest score accepted?
Solution
1.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 71
u = mean = 80
s = standard deviation = 6
Thus,
z = (x - u) / s = -1.5
Thus, using a table/technology, the right tailed area of this is
P(z > -1.5 ) = 0.933192799 [ANSWER]
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2.
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 89
x2 = upper bound = 92
u = mean = 80
s = standard deviation = 6
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 1.5
z2 = upper z score = (x2 - u) / s = 2
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.933192799
P(z < z2) = 0.977249868
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.044057069 [ANSWR, BETWEEN 89 AND 92]
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3.
First, we get the z score from the given left tailed area. As
Left tailed area = 0.975
Then, using table or technology,
z = 1.959963985
As x = u + z * s / sqrt(n)
where
u = mean = 80
z = the critical z score = 1.959963985
s = standard deviation = 6
Then
x = critical value = 91.75978391 [lowest score with merit]
