A normal distribution has = 20 and = 4. What is the probability of randomly sele
ID: 3126564 • Letter: A
Question
A normal distribution has = 20 and = 4. What is the probability of randomly selecting a score greater than 25 from this distribution?
E. None of the above.
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A normal distribution has a mean of µ = 40 with = 8. If one score is randomly selected from this distribution,
what is the probability that the score will be less than X = 34?
D. 0.4532 or 45.32%
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A normal distribution has a mean of µ = 80 with = 20. What score separates the highest 15% of the distribution from the rest of the scores? (Note: use the closest value listed in the Unite Normal table).
A. 0.3944 or 39.44% B. 0.1056 or 10.56% C. 0.8944 or 89.44% D. 0.7888 or 78.88%E. None of the above.
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A normal distribution has a mean of µ = 40 with = 8. If one score is randomly selected from this distribution,
what is the probability that the score will be less than X = 34?
A. 0.7734 or 77.34% B. 02266 or 22.66% C. 0.2734 or 27.34%D. 0.4532 or 45.32%
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A normal distribution has a mean of µ = 80 with = 20. What score separates the highest 15% of the distribution from the rest of the scores? (Note: use the closest value listed in the Unite Normal table).
A. X = 59.2 B. X = 100.8 C. X = 95 D. X = 65Explanation / Answer
1.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 25
u = mean = 20
s = standard deviation = 4
Thus,
z = (x - u) / s = 1.25
Thus, using a table/technology, the right tailed area of this is
P(z > 1.25 ) = 0.105649774 [ANSWER, B]
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2.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 34
u = mean = 40
s = standard deviation = 8
Thus,
z = (x - u) / s = -0.75
Thus, using a table/technology, the left tailed area of this is
P(z < -0.75 ) = 0.226627352 [ANSWER, B]
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3.
First, we get the z score from the given left tailed area. As
Left tailed area = 0.85
Then, using table or technology,
z = 1.036433389
As x = u + z * s,
where
u = mean = 80
z = the critical z score = 1.036433389
s = standard deviation = 20
Then
x = critical value = 100.728667-->100.8 [closest, OPTION B]
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