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What is the relative frequency of score less than 120 using the 68-95-99.7 rule if a set of test scores is normally distributed with a mean of 100 and a standard deviation of 20?
Wiki User
∙ 10y ago
120 is one standard deviation greater than the mean (z = 1).
So you want Pr(z < 1)
The probability of a score at most 1 sd away from the mean is 0.68
That is Pr(-1 < z < 1) = 0.68
So Pr(|z| > 1) = 1 - 0.68 = 0.32
then by symmetry,
Pr(z > 1) = 1/2*0.32 = 0.16
So Pr(z < 1) = 1 - 0.16 = 0.84
120 is one standard deviation greater than the mean (z = 1).
So you want Pr(z < 1)
The probability of a score at most 1 sd away from the mean is 0.68
That is Pr(-1 < z < 1) = 0.68
So Pr(|z| > 1) = 1 - 0.68 = 0.32
then by symmetry,
Pr(z > 1) = 1/2*0.32 = 0.16
So Pr(z < 1) = 1 - 0.16 = 0.84
120 is one standard deviation greater than the mean (z = 1).
So you want Pr(z < 1)
The probability of a score at most 1 sd away from the mean is 0.68
That is Pr(-1 < z < 1) = 0.68
So Pr(|z| > 1) = 1 - 0.68 = 0.32
then by symmetry,
Pr(z > 1) = 1/2*0.32 = 0.16
So Pr(z < 1) = 1 - 0.16 = 0.84
120 is one standard deviation greater than the mean (z = 1).
So you want Pr(z < 1)
The probability of a score at most 1 sd away from the mean is 0.68
That is Pr(-1 < z < 1) = 0.68
So Pr(|z| > 1) = 1 - 0.68 = 0.32
then by symmetry,
Pr(z > 1) = 1/2*0.32 = 0.16
So Pr(z < 1) = 1 - 0.16 = 0.84
Wiki User
∙ 10y ago
Wiki User
∙ 10y ago120 is one standard deviation greater than the mean (z = 1).
So you want Pr(z < 1)
The probability of a score at most 1 sd away from the mean is 0.68
That is Pr(-1 < z < 1) = 0.68
So Pr(|z| > 1) = 1 - 0.68 = 0.32
then by symmetry,
Pr(z > 1) = 1/2*0.32 = 0.16
So Pr(z < 1) = 1 - 0.16 = 0.84
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What is the Difference in relative frequency and percentage frequency?
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The first class in a relative frequency table is 50-59 and the corresponding relative frequency is 0.2What does the 0.2 value indicate?
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Relative frequency formula?
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