In: Statistics and Probability
A teacher looks at the scores on a standardized test, where the mean of the test was a 73 and the standard deviation was 9. Find the following
z-score table Link
Given :
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The probability that a student will score between 73 and 90 = P(73<x<90)
First find the z-scores for x=73 and x=90
For x=73
For x=90
Required probability is P( 0 < z < 1.89 )
= P( z < 1.89 ) - P( z < 0 )
= 0.9706 - 0.5000 ( From the z-score table for z=1.89 and for z =0)
= 0.4706
The probability that a student will score between 73 and 90 is 0.4706
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the probability that a student will score between 65 and 85 = P(65<x<85)
First find the z-scores for x=65 and x=85
For x=65
For x=85
Required probability is P( -0.89 < z < 1.33 )
= P( z < 1.33 ) - P( z < -0.89 )
= 0.9082 - 0.1867 ( From the z-score table for z=1.33 and for z = -0.89 )
= 0.7215
The probability that a student will score between 73 and 90 is 0.7215
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the probability that a student will get a score greater than 70 = P(x>70)
using the z-score formula
here x=70
Required probability is P(z>-0.33) = P(z<0.33)
= 0.6293 ( From z-score table for z=0.33)
The probability that a student will get a score greater than 70 is 0.6293
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the probability that a student will score more than 80 = P(x>80)
using the z-score formula
here x=80
Required probability is P(z>0.78) = 1 - P(z<0.78)
= 1 - 0.7823 ( From z-score table for z=0.78 )
= 0.2177
The probability that a student will score more than 80 is 0.2177
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score would give you the top 12.1%
P(Z>z) = 0.1210
P(Z<z) = 1 - 0.1210
P(Z<z) = 0.8790
Now looking in body (middle part ) of z-score table for area closest to 0.8790 then note down the corresponding z value.
z = 1.17
Now use z-score formula to get the value of x ( required score)
The score that would give you the top 12.1% is 83.53
Rounded to nearest whole number would be 84