Section 8.5 The Normal Distribution

The Normal random variable X = x is a random variable. It’s defined by the mean

µ and standard deviation � of the normal distribution.

Properties of the Normal Distribution Curve

1. The curve has a peak at x = µ.

2. The curve is symmetric about the line x = µ.

3. The curve always lies above the x-axis but approaches the x-axis as x extends indefinitely in either

direction.

4. The total area under the curve is 1.

5. The Standard Normal distribution has µ = 0 and � = 1. We denote the standard normal

random variable with Z = z.

Calculating the Probability of a Normal Random Variable

The probability P (a < X < b) that X lies between a and b is the area under the curve between x = a

and x = b. This can be found using probability tables but in this class we will use the calculator

function “normalcdf” to calculate probabilities for a Normal random variable.

-

→

k

oh .

continuous-

- =-

- -

-

Total probability for thee distribution

- - -

-

-

-

I-

pcacxcb )

Calculator Steps:

Click 2ND , VARS , 2 . You should see normalcdf( on your screen.

The format is normalcdf(smallest x-value/z-value, biggest x-value/z-value, µ, �) . Use E99 if the

biggest x-value/z-value is 1 and �E99 if the smallest x-value/z-value is �1. To get E99 click

2ND , , .

Note: We never use “normalpdf” in this class.

1. Answer the following:

(a) Sketch the area under the standard normal curve corresponding to P (�0.29 < Z < 0.29),

and find the probability. Round to 4 decimal places.

(b) Sketch the area under the standard normal curve corresponding to P (Z < �0.29), and find

the probability. Round to 4 decimal places.

(c) Sketch the area under the standard normal curve corresponding to P (Z > 0.29), and find

the probability. Round to 4 decimal places.

2 Fall 2019, Maya Johnson

.-

-

t

Tf ?% PC - 0.294×20.29#✓\Normaledff -

.29

,-29,0 ,

t )

T.io = 0.2282J

*

*

" it " " " = N'

¥-9:* ,

-

--0.3859J

-

f)egad plz > 0.29 )e

'

¥\ Norouolcdfl. 29,5-990,1 )

a =as

2. SupposeX is a normal random variable with µ = 376 and � = 17. Find the following probabilities.

(Give answers to four decimal places.)

(a) P (X < 409)

(b) P (389 < X < 411)

(c) P (X > 409)

Inverse Normal Distribution:

Suppose we are given the probability or area under the curve and are asked to find the random

variable value that corresponds to the given probability. To solve this problem we will use the

calculator function “invNorm.”

Calculator Steps:

Click 2ND , VARS , 3 . You should see invNorm( on your screen.

The format is invNorm(probability to the left of X = x or Z = z, µ, �) .

3 Fall 2019, Maya Johnson

- -

= Normal cdf ( - E 99, 409 , 376 ,

17 )

1¥

=a④¥1326409

= Normale df ( 389 ,44 ,

376 ,17 )

T.in=

= Normal edf ( 409 ,E 99

,376,17 )

.

=

- -

=

-- - -

T 99

PlXI a 5

3. Let Z be the standard normal variable. Find the values of a that satisfy the given probabilities.

(Give answers to four decimal places.)

(a) P (Z > a) = 0.2829

(b) P (Z < a) = 0.2441

(c) P (�a < Z < a) = 0.4441

4 Fall 2019, Maya Johnson

-

-

Az luv Norm ( I -

. 282920,1 )

I

""

=o.sr value

→ a *

- * →

¥ right

a =Inn Norml . 2441,011 )

a .o

- ea neo

←left

1--

A ① pro

←- a = inv Norm (1-24441-2011)

left

x ta 4441 tx = I - 0.5889

2x to 4441=1

212--1-24-441

⇒a-ao.SE#yx=I-.444t=

.

27795

2

4. Find the indicated quantities given that X is a normal random variable with a mean of 40 and a

standard deviation of 10. (Round answers to four decimal places.)

(a) Find the value of b such that P (X b) = 0.1515

(b) Find the value of c such that P (X � c) = 0.9678

(c) Find the values of A and B such that P (A X B) = 0.9146 if A and B are symmetric

about the mean.

5 Fall 2019, Maya Johnson

- pea40¥

-

b z inv Norm ( O .1515,40 ,

10 )

\ = 29.6998J

c-

left

H-

-

c = in Norm ( I -

.

9678,4910 )qb

= 21.50600

1-c I

9ftright

-- -

-

.

B

A =inv Norm ( no 427 ,

40 ,to )

= 22.7982J

X -

- 1-092146-2.0427 B = in Norml .0427 t

.

9146,40,

10 )

= 57.2018J

Section 8.6 Applications of the Normal Distribution

1. On the average, a student takes 119 words/minute midway through an advanced court reporting

course at the American Institute of Court Reporting. Assuming that the dictation speeds of the

students are normally distributed and that the standard deviation is 1, 440 words/hour, find the

probability that a student randomly selected from the course can make dictation at the following

speeds. (Give answers to four decimal places.)

(a) more than 167 words/minute

(b) less than 71 words/minute

2. The weight of topsoil sold in a week is normally distributed with a mean of 800 tons and a

standard deviation of 32 tons. (Round answers to two decimal places.)

(a) What percentage of weeks will sales exceed 864 tons?

(b) What percentage of weeks will sales be between 752 and 816 tons?

6 Fall 2019, Maya Johnson

mean

- -

- -

11440 wordsa = n-y.FI#m.Tote

-⇒ 6=24 words turn .

PCX 7167 )

= Normal cdf ( 167 ,E 99

, 119,24 )

= 0.0228J

-

PC XL 71 )

= Normal edf ( - E 99,

71,

119,

24 )

= 0.6228J These two are symmetricabout

the mean

-

- After plugins by 100

PIX > 00%-

= Normal cdf I 864,

E 99,800,32 ) *100%=2.289€-

pl 752 LX L 816 ) x

100%=

' Normale If ( 752,816,800,323×100 %

= 62.470€

3. A teacher wishes to ”curve” a test whose grades were normally distributed with a mean of 60 and

standard deviation of 15. The top 10% of the class will get an A, the next 30% of the class will

get a B, the next 35% of the class will get a C, the next 20% of the class will get a D and the

bottom 5% of the class will get an F . Find the cuto↵ for each of these grades. (Round answers

to two decimal places.)

(a) The A cuto↵ is a grade of

(b) The B cuto↵ is a grade of

(c) The C cuto↵ is a grade of

(d) The D cuto↵ is a grade of

7 Fall 2019, Maya Johnson

--

- --

-- -

IDB

inv Norm ( l -

.10

,60 ,

15 )

=

79.222Jinv norm ( I -

. 4,

60,

I 5)

= 63.80J

innocent I -

.75

,60

, 15 )= 49.88J

inv Norm fl -

.95

,60

,15 )

= 35.33J

4. The distribution of heights of adult males is normally distributed with mean 63 inches and stan-

dard deviation 2.4 inches. Answer the following. (Round answers to 2 decimal places.)

(a) What minimum height is taller than 64% of all adult males?

(b) What two heights that are symmetric about the mean enclose the middle 82% of all heights?

8 Fall 2019, Maya Johnson

- -

-

-

= in , Noam C. 64,

63,2 . 4).

-

= 63.86inch

-c-

Shorter

-

P ( A LX L B) =.

82

.

X -

-

1-2-82= .

09

A = inv Norm ( .09

,63 ,

2.4 )

= -9.78inchB = inv Norm ( .

09 t .82 ,63 ,

2. 4)

a 66.22inch