Question

**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?

Answer #1

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**

-------------------

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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