Question

# 1) Let y be a random variable having a normal distribution with a mean equal to...

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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