Transcript Chapter 7

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
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:
• z =(x1 -µ)/δ= 19-21/1=-2.00
•z =(x 2-µ)/δ=21-21/1=0
• From table the area to the left of z 1=-2.00 is 0.0228.
• Area to left of z 2 =0 is 0.5000
• P(19<x<21) = 0.5000-0.0228= 0.4772
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