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**Standard Normal Distribution**

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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 36.5 42.7 40.6 47.5 48.4 37.5 45.5 41.2 40.5 44.4 42.6 42.0 40.3 38.5 43.6 45.0 40.7 36.3 37.6 48.5 41.7 38.9 42.3 38.8 41.9 42.1 44.6 37.4 38.2 44.0 35.9 43.7 48.1 38.7 46.0 37.7 34.6 47.0 42.8 39.9

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**Construct a relative frequency histogram**

Draw a normal density curve on relative frequency histogram

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**Construct a relative frequency histogram**

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**Construct a relative frequency histogram**

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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 The area to the left of z=-0.46 is The area to the right of -0.46= 0.6772

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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) =0.8893

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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 which corresponds to z-score of -0.47

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Question 4 Find the z-score so that the area to the right of the z-score is Solution: Find the area to the left of the unknown z-score Area to the left=1-area to the right 0.5668 From the table find an area closest to Area closest to is Corresponding z-score of is 0.17

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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 = Area to the right = 0.9998

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**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 = Area to the right= 0.7088

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**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 = Area to the right = 0.0129

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**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 = Area to the right = 0.0003

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Question 6 The mean incubation time of fertilized chicken eggs kept at 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

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**The mean incubation time of fertilized chicken eggs kept at 100**

The mean incubation time of fertilized chicken eggs kept at 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 Area to left of z=1.00= =0.1587

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**The mean incubation time of fertilized chicken eggs kept at 100**

The mean incubation time of fertilized chicken eggs kept at 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 Area to left of =0 is P(19<x<21) = =

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**CSTEM Web link http://www.cis.famu.edu/~cdellor/math/**

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Copyright Tim Morris/St Stephen's School

Copyright Tim Morris/St Stephen's School

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