Statistics assignment 3
W3A3 - Statistic_problems/normalppt.pptx
PAGE 1: X<0
PAGE 2: X>0
The area to the left of z=1.36 is 0.9131.
That means the probability of choosing
a standard normal variable less than 1.36
is a little over 91%.
.9131
To find the probability that z falls below a certain value k, simply
look up k on the table.
Example:
P(z<-0.6)=
0.2743
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To find the probability that z falls above a certain value k, we
do one of two things. We can:
a) look up k on the table, and subtract that number from 1,
since P(z>k) = 1 - P(z<k)
b) look up -k, and note that P(z>k) = P(z<-k), since the area
is symmetric about zero.
Example:
P(z>0.38)=
1 - 0.6480 = 0.352
= P(z<-0.38)
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To find the probability that z falls between values j and k, we
note that P(j<z<k) = P(z<k) - P(z<j). Thus we can just look up
k and j, and subtract the smaller area from the larger one.
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EXAMPLE:
P(-0.89 < z < 1.44) =
P(z< 1.44) - P( z <-0.89 ) =
0.9251 - 0.1867 =
0.7384
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Often we want to find the z-value or values that cut off a
specific percentage of data. For example, we can ask:
What z-scores bound:
the bottom 10% ?
the top 12% ?
the middle 30% ?
To answer these questions we must look inside the table to find
the required percentage, and match that with the z-score.
So, to find the cutoff for the bottom 10%, we look in the table for
the closest value to 0.1:
The closest we can get is 0.1003, which corresponds to z = -1.28.
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To find the top 12%, we note that the same number will cut
off the bottom 88%, since 100 - 12 = 88.
The closest we can get to 0.88 is:
0.8810
The z-score that cuts off the top 12% is then z=1.18.
To find the middle 30%, we note that 100 - 30 = 70, so the
remaining 70% must be split evenly below and above the middle
30% The closest we can get to 0.35 is:
The z-scores that cut off the middle 30% are then z = + 0.39.
0.3483
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We can combine this information with the z-score formula now to
answer questions about any normal distribution, not just N(0,1).
Consider your SAT score. The verbal and math scores are
designed to fall between 200 and 800 in a roughly normal
distribution with mean 500 and SD 100: N(500,100).
What % of students score above 670?
z = 670 - 500 = 1.7
100
1 - 0.9554 = 0.0446.
Only about 4.5%.
P(z>1.7) = 1 - P(z<1.7)=
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What % of students score between 430 and 530?
530: z = 530 - 500 = 0.3
100
0.6179 - 0.2420 = 0.3759.
About 37.6%.
430: z = 430 - 500 = -0.7
100
P(-0.7<z<0.3) = P(z<0.3) - P(z<-0.7)=
What SAT score puts a student in the top 1%?
The closest we can get to 0.99 is: 0.9901
z = 2.33 means:
X = (2.33)(100) + 500 = 233 + 500 = 733
A score of 740 or higher puts you in the top 1%.
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