Question

Scores on a test have a mean of 71 and Q3 is 82. The scores have a distribution that is approximately normal. Find P90

Answer #1

**Given:**

Mean = 71

Q3 = 75th percentile = 82

We also know that

From the standard normal table or using Excel the corresponding z-value for 0.75(Q3 or 75th percentile) is calculated as 0.6745. (In Excel, use the function NORMSINV(0.75))

Therefore, the above equation of Z can be re-written for the value of 82 (Q3) as

Now, solving the above equation we have

Now, we have found standard deviation .

From the standard normal table or using Excel we can find the Z-value corresponding to 90th percentile as 1.2816 (In Excel, use the function NORMSINV(0.90)).

Hence, 90th percentile value (value of x in the Z formula) is calculated using the above z-formula as below

Hence, **P90 = 92 (approx).**

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Scores on a test have a mean of 72.9 and 9 percent of the scores
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On a math test, the
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98 71 75 65
75 75 98 82
75 66 83 71
71 83 75 71
83 75 71 82
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Construct a frequency
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Fill out in the blank
below
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Frequency

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Group of answer choices
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73 and 91
76 and 88
64 and 100

forty students took a test on which the mean score was an 82 and
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A population of scores forms a normal distribution with a mean
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places.)
(b) If samples of size n = 15 are selected from the
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