Standard Normal Distribution

The following frequency distribution represents the heights (in inches) of eighty randomly selected 5-year olds.

44.5 42.4 42.2 46.2 45.7 44.8 43.3 39.5

45.4 43.0 43.4 44.7 38.6 41.6 50.2 46.9

39.6 44.7 36.5 42.7 40.6 47.5 48.4 37.5

45.5 43.3 41.2 40.5 44.4 42.6 42.0 40.3

42.0 42.2 38.5 43.6 40.6 45.0 40.7 36.3

44.5 37.6 42.2 40.3 48.5 41.6 41.7 38.9

39.5 43.6 42.3 38.8 41.9 40.3 42.1 41.9

42.3 44.6 40.5 37.4 44.5 40.7 38.2 42.6

44.0 35.9 43.7 48.1 38.7 46.0 43.4 44.6

37.7 34.6 42.4 42.7 47.0 42.8 39.9 42.3

(a) Construct a relative frequency histogram

(a) Draw a normal density curve on relative frequency histogram

(a) Construct a relative frequency histogram

(a) Construct a relative frequency histogram

Microsoft Office Excel 97-2003 Worksheet

Question 1

• Find the area under the standard normal curve to the right of z= -0.46

• Solution:– Find the row that represents -0.4 and the column

that represents 0.06. The area to the left of z=-0.46 is 0.3228

– The area to the right of -0.46=1-0.3228

0.6772

Question 2

• Find the area under the standard normal curve between z=-1.35 and z=2.01

• Solution:– Find the area to the left of z=2.01– Find the area to the left of z=-1.35– The area under the standard normal curve between

z=2.01 and z=1.35 is:– (area to left of z=2.01)-(area to right of z=-1.35)– 0.9778-0.0885=0.8893

Question 3

• Find the z-score so that the area to the left of the z-score is 0.32

• Solution:– Look for area in the table closest to 0.32– Find z-score that corresponds to the area closest to

0.32– From table closest area of 0.32 is 0.3192 which

corresponds to z-score of -0.47

Question 4

• Find the z-score so that the area to the right of the z-score is 0.4332

• Solution:– Find the area to the left of the unknown z-score– Area to the left=1-area to the right– 1-0.4332– 0.5668– From the table find an area closest to 0.5668– Area closest to 0.5668 is 0.5675– Corresponding z-score of 0.5668 is 0.17

Question 5

• Determine the area under the standard normal curve that lies to the right of:

(a) z=-3.49– Solution:– Area to the right=1-area to the left– Area to the left = 0.0002– Area to the right =1-0.0002– 0.9998

• Determine the area under the standard normal curve that lies to the right of:

(b) z=-0.55• Solution:

– Area to the right=1-area to the left– Area to the left = 0.2912– Area to the right=1-0.2912– 0.7088

• Determine the area under the standard normal curve that lies to the right of:

(c) z=-2.23• Solution:

– Area to the right=1-area to the left– Area to the left = 0.9871– Area to the right =1-0.9871– 0.0129

• Determine the area under the standard normal curve that lies to the right of:

(d) z=3.45• Solution:

– Area to the right=1-area to the left– Area to the left = 0.9997– Area to the right =1-0.9997– 0.0003

Question 6• The mean incubation time of fertilized chicken

eggs kept at 100.5 F in a still air incubator is 21 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 1 day.

(a)What is the probability that a randomly selected chicken hatches in 20 days

• Solution:– Z=(x-µ)/δ= 20-21/1=-1.00– From table the area to the left of z=-1.00 is 0.1587

• The mean incubation time of fertilized chicken eggs kept at 100.5 F in a still air incubator is 21 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 1 day.

(b)What is the probability that a randomly selected chicken egg takes over 22 eggs to hatch

• Solution:– Z=(x-µ)/δ= 22-21/1=1.00– From table the area to the left of z=1.00 is 0.8413.– Area to left of z=1.00=1-0.8413=0.1587

• The mean incubation time of fertilized chicken eggs kept at 100.5 F in a still air incubator is 21 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 1 day.

(c)What is the probability that a randomly selected chicken egg hatches between 19 and 21 days

• Solution:• =( -µ)/δ= 19-21/1=-2.00• =( -µ)/δ=21-21/1=0• From table the area to the left of =-2.00 is 0.0228.• Area to left of =0 is 0.5000• P(19<x<21) = 0.5000-0.0228= 0.4772

1z

1x

2z

2x1z

2z

CSTEM Web link

http://www.cis.famu.edu/~cdellor/math/

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