1) Let y be a random variable having a normal
distribution with a mean equal to 50
and a standard deviation equal to 8. Find the following
probabilities
a) P(|y| > 63)
b) P(|y| > 23)
2) The College Boards, which are administered
each year to many thousands of
high school students, are scored so as to yield a mean of 500 and a
standard
deviation of 100. These scores are close to being normally
distributed. What
percentage of the scores can be expected to satisfy each
condition?
a) Greater than 650
b) Greater than 750
c) What is the 90th percentile?
3) The ACME Stinker was an automobile that is
no longer in production. When
driven in city traffic, the nitrogen oxide levels it emitted
adhered approximately to
a normal distribution with mean 2.1 grams per mile and standard
deviation 0.3
grams per mile.
a) By EPA mandate, nitrogen oxide levels were not
to exceed 2.7 grams per
mile. About what proportion of Stinkers were in violation?
b) At most 25% of Stinkers exceeded what nitrogen
oxide level?
c) ACME was forced to reduce the nitrogen oxide
emissions level so that no
more than 1% of Stinkers exceeded 2.7 grams per mile. If the
standard
deviation remained at 0.3 grams per mile, to what value did the
mean
emissions level have to be reduced?
1). Population mean, µ = 50
Population standard deviation, σ = 8
a) P(y > 63) =
= P( (y-µ)/σ > (63-50)/8 )
= P(z > 1.625)
= 1 - P(z < 1.625)
Using excel function:
= 1 - NORM.S.DIST(1.625, 1)
= 0.0521
b) P(y > 23) =
= P( (y-µ)/σ > (23-50)/8 )
= P(z > -3.375)
= 1 - P(z < -3.375)
Using excel function:
= 1 - NORM.S.DIST(-3.375, 1)
= 0.9996
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2). Population mean, µ = 500
Population standard deviation, σ = 100
a) P(X > 650) =
= P( (X-µ)/σ > (650-500)/100)
= P(z > 1.5)
= 1 - P(z < 1.5)
Using excel function:
= 1 - NORM.S.DIST(1.5, 1)
= 0.0668
b) P(X > 750) =
= P( (X-µ)/σ > (750-500)/100)
= P(z > 2.5)
= 1 - P(z < 2.5)
Using excel function:
= 1 - NORM.S.DIST(2.5, 1)
= 0.0062
c) Z score at p = 0.90 using excel = NORM.S.INV(0.9) = 1.2816
Value of X = µ + z*σ = 500 + (1.2816)*100 = 628.1552
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