Upon reaching orbit the Space Shuttle will engage in MECO (Main Engine Cut-Off). At that time it will be traveling at an average of 17,580mph (σ = 35mph). A speed between 17500 and 17660 is needed to keep the Shuttle in orbit. Below 17500 is too slow. a.) What proportion of Shuttles reach this target interval?

b.) What proportion of Shuttles have to fire secondary engines to correct speed because they are going to slow?c.) The top 1% of Shuttles accelerate over what speed?

P(17500 < x < 17660) = P(17500 < x < 17660) = Normalcdf(-2.286, 2.286)Normalcdf(-2.286, 2.286)

a.) 0.978a.) 0.978

b.) 0.011b.) 0.011

c.) 17661.4 mphc.) 17661.4 mph

P(x < 17500) = P(x < 17500) = Normalcdf(-E99, -2.286)Normalcdf(-E99, -2.286)

InvNorm(.99) = 2.326InvNorm(.99) = 2.3262.326 = (X – 17580)/352.326 = (X – 17580)/35

WARM – UP Jake receives an 88% on a Math test which had a N(81, 8). That same day he takes an English test N(81, 12) and receives a 90%. Which test did he do better on with respect to the rest of the class? Support your answer by converting each score into a z-score and then calculating the percent of students he did better than for each test (percentile = P(x<#)P(x<#).

Math: P(x < 88) =Math: P(x < 88) =

English: P(x < 90) =English: P(x < 90) =

P(z < 0.875) =P(z < 0.875) =

z x

88 81

8z

Normalcdf(-E99, 0.875)Normalcdf(-E99, 0.875)

Jake did better than Jake did better than 80.9% of his peers.80.9% of his peers.

P(z < 0.75) =P(z < 0.75) = Normalcdf(-E99, 0.75)Normalcdf(-E99, 0.75)

Jake did better than Jake did better than 77.3% of his peers.77.3% of his peers.

90 8112

z

Normal Proportion Calculations● State the Problems as an Inequality:

P(x < #) or P(x > #) or P(a < x < b)● Draw a picture if necessary.

● Standardize the value by using the z-score formula.

● Use the Calculator [normalcdf (lower, upper)] to find the proportion.

● Draw a picture if necessary.● Find the Standardized value (z) by using [InvNorm(.%)].

(% to the LEFT)● Solve for the x within the z-score formula.

Finding Values from Proportions

z x

1. A recent Statistics Test produced a .What proportion of the class scored below a 75 on the test?

86,6N

75P x

P( x < 75 ) =P( x < 75 ) =

z x

866

75z P(z < -1.833)P(z < -1.833)

normalcdf(-E99, -1.833) =normalcdf(-E99, -1.833) =

68 74 80 86 92 98 104 -3 -2 -1 0 1 2 3

First, Convert 75 to a z-score

0.03340.0334

Normcdf = Proportion or %

InvNorm = z-score

2. A recent Statistics Test produced a .

What score did a statistics student need to have earned in order to be within the Top 10% of the class?

Hint: a) First find the z-score for the 90th Percentile.

b) Within the z-score formula substitute the z, the μ, the σ, and then solve for x.

861.28166x

X X = 93.7 = 94%= 93.7 = 94%

InvNorm(.90) = 1.2816InvNorm(.90) = 1.2816

10%90%

z x

86,6N

3. Between what scores does a student have to get to be in the Middle 50% of the class?

860.6746x

81.956 < X < 90.04481.956 < X < 90.044

InvNorm(.25) = -0.674InvNorm(.25) = -0.674

860.6746x

Q1 Q3

InvNorm(.75) = 0.674InvNorm(.75) = 0.674

86,6N

50%

4. The ACT exam scores follow a normal distribution. If the top 10% scored above a 34 and the bottom 20% scored below a 18, what is the mean and

standard deviation?Hint: a) First find the z-score for the 90th Percentile.

Then find the z-score for the 20th Percentileb) Within two z-score formulas substitute the z and

x and then solve for the μ and the σ.

1.282 34

InvNorm(.90) = 1.282InvNorm(.90) = 1.28218.842

InvNorm(.20) = -0.842InvNorm(.20) = -0.842

1.282 34 .842 18 1.282 34

.842 18

ApplicationA headache medicine is effective if it demonstrates relief (a score of 5 or higher) for over 75% of the population. If patients are asked to describe the effects of a new drug by rating it on a scale from 1(NO Effect) to 10(Complete Relief), and their responses follow a N(5.8, 1.3):

1. What Proportion of the population indicated a relief score of 3 or below?

2. What score represents the top 75% of the population?

3. What Proportion of the population indicate a relief score 5 or higher?

X: X: 1.9 3.2 4.5 5.8 7.1 8.4 9.71.9 3.2 4.5 5.8 7.1 8.4 9.7Z: -3 -2 -1 0 1 2 3Z: -3 -2 -1 0 1 2 3

3 5.83 2.151.3

( 99, 2.15) 0.0156

P x z

normalcdf E

0.0156

InvNormx x

(. ) .

. ..

.

25 67

67 5813

4 923

75%

4.923

5 5.85 0.621.3

( 0.62, 99) 0.7308

P x z

normalcdf E

73.08%z x

1. If the GPA of seniors follows ,what proportion of the senior class has a GPA

below 3.0?

2.48,0.53N

3.0 ?P x

P( x < 3.0 ) =P( x < 3.0 ) =

z x

3.0 2.480.53

z P(z < 0.9811)P(z < 0.9811)

normalcdf(-E99, 0.9811) =normalcdf(-E99, 0.9811) = 0.83670.8367

-3 -2 -1 0 1 2 3-3 -2 -1 0 1 2 3

2. If the GPA of seniors follows ,what GPA would represent the top 5% of the

senior class?

2.481.64490.53x

X X = = 3.3523.352

InvNorm(.95) = 1.6449InvNorm(.95) = 1.6449

90%

z x

2.48,0.53N

Hint: a) First find the z-score for the 95th Percentile.b) Within the z-score formula substitute the z95, the μ, and the σ, and then solve for x. -3 -2 -1 0 1 2 3-3 -2 -1 0 1 2 3

? .05P x

0.050.05

Between what GPA’s does a student have to get to be in the Middle 50% of the senior class if the distribution is ?

2.480.6740.53x

2.123 < X < 2.8372.123 < X < 2.837

InvNorm(.25) = -0.674InvNorm(.25) = -0.674

2.480.6740.53x

Q1 Q3

50%50%25%25% 25%25%

2.48,0.53N

InvNorm(.75) = 0.674InvNorm(.75) = 0.674

z x

IQR = Q3 – Q1

You are interested in finding the proportions of high school students who travel various miles for college. An approximately normal distribution of students reveal a N(210, 88).

1) What proportion traveled more than 300 miles?

2) What proportion traveled more than 450 miles?

3) What proportion traveled less than 100 miles?

4) What distance, in miles, represents the 90th Percentile?

5) What distance, in miles, do the top 5% represent?

6) What percent of students stay close to home by traveling within 5 to 60 miles?

0.15320.1532

0.003190.00319

322.8 miles322.8 miles

355 miles355 miles

3.42%3.42%

0.10560.1056

P(x > 300) =P(x > 300) = P(z > 1.0227) P(z > 1.0227) Normalcdf(1.0227, E99) Normalcdf(1.0227, E99)

z x

P(x > 450) =P(x > 450) = P(z > 2.7273) P(z > 2.7273) Normalcdf(2.7273, E99) Normalcdf(2.7273, E99)

P(x < 100) =P(x < 100) = P(z < -1.25) P(z < -1.25) Normalcdf(-E99, -1.25) Normalcdf(-E99, -1.25)

InvNorm(.90)InvNorm(.90) z = 1.2816z = 1.2816 1.2816 = (X – 210)/881.2816 = (X – 210)/88

InvNorm(.95)InvNorm(.95) z = 1.6449z = 1.6449 1.6449 = (X – 210)/881.6449 = (X – 210)/88

P(5 < x < 60) =P(5 < x < 60) =P(-2.3295 < z < -1.7045)P(-2.3295 < z < -1.7045) Normalcdf(-2.3295, -1.7045) Normalcdf(-2.3295, -1.7045)