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5.3 and 5.4 Standard Deviation & Normal Distribution (Solutions).notebook December 14, 2015 5.3 & 5.4 Standard Deviation and Normal Distribution Standard deviation: A measure of how scattered the data values are in relation to the mean. It's always positive. low standard deviation = most data values are close to the mean. high standard deviation = data values are scattered farther from the mean. The standard deviation, σ , can be expressed using symbols: σ = Ʃ(x - μ ) 2 N N = Population (# of data pieces) x = each piece of data μ = mean The symbol Ʃ means summation (total)

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Page 1: 5.3 and 5.4 - Standard Deviation & Normal Distribution ...killornmath.weebly.com/uploads/3/0/9/...standard...normal_distributio… · 5.3 and 5.4 Standard Deviation & Normal Distribution

5.3 and 5.4 ­ Standard Deviation & Normal Distribution (Solutions).notebookDecember 14, 2015

5.3 & 5.4

Standard Deviation

and

Normal Distribution

Standard deviation: A measure of how scattered the data values are in relation to the mean.  

It's always positive.

low standard deviation = most data values are close to the mean.  

high standard deviation = data values are scattered farther from the mean.

The standard deviation, σ, can be expressed using symbols:

σ = Ʃ(x - μ )2N√

N = Population (# of data pieces)

x = each piece of data

μ = meanThe symbol Ʃ means summation (total)

Page 2: 5.3 and 5.4 - Standard Deviation & Normal Distribution ...killornmath.weebly.com/uploads/3/0/9/...standard...normal_distributio… · 5.3 and 5.4 Standard Deviation & Normal Distribution

5.3 and 5.4 ­ Standard Deviation & Normal Distribution (Solutions).notebookDecember 14, 2015

This looks complicated but it really means that we...1) Calculate the mean (μ) 2) Find the difference between each of the data values and 

the mean.  3) Square all of the differences and add them together.  4) Divide this sum by the number of pieces of data.  5) Take the square root of that number, and it gives the 

STANDARD DEVIATION.

Example #1: Find the mean and standard deviation for the prices (in dollars) of cordless phones at an electronics store, to the nearest tenth. 

45, 50, 60, 60, 65, 80 

Item Number Mean Deviation Deviation2

Totals

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5.3 and 5.4 ­ Standard Deviation & Normal Distribution (Solutions).notebookDecember 14, 2015

Example #2:  Calculate the standard deviation of the following test data to the nearest hundredth.

Test Scores:  22, 99, 33, 57, 75, 81, 62, 29

Item Number Mean Deviation Deviation2

1 22 57.25 1242.56

2 99 57.25 41.75 1743.06

3 33 57.25 ­24.25 588.064 57 57.25 ­0.25 0.06255 75 57.25 17.75 315.066 81 57.25 23.75 564.067 62 57.25 4.75 22.568 29 57.25 ­28.25 798.06   Totals 458 458 0 5273.48

­35.25

p. 233

# 2, 6

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5.3 and 5.4 ­ Standard Deviation & Normal Distribution (Solutions).notebookDecember 14, 2015

Frequency distributions can have different shapes.

• Sometimes the distribution of data has a  special shape. 

For example, the purple graph has one peak, so the shape has one mode. 

Describe the shape of each graph, and suggest a context that the graph could represent.

Normal Distribution

When data is normally distributed in a bell shape known as the normal curve.

It represents the data with a mean"μ"and standard deviation "σ"

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5.3 and 5.4 ­ Standard Deviation & Normal Distribution (Solutions).notebookDecember 14, 2015

The properties of a normal curve can be summarized as follows:

• The normal curve is shaped like a bell, peaking in the middle at the mean, sloping down toward the sides, and approaching zero at the extremes.

• About 68% of the data is within 1 standard deviation of the mean.

• About 95% of the data is within 2 standard deviations of the mean.

• About 99.7% of the data is within 3 standard deviations of the mean.

• The area under the curve can be considered 1 unit, since it represents 100% of the data.

• Generally, measurements of living things (such as mass, height, and length) have a normal distribution.

μ μ + σμ - σμ - 2σ μ + 2σμ - 3σ μ + 3σ

95%68%

99.7%

2.35%13.5%13.5%

34%34%

2.35%

THE NORMAL CURVE

34%34%

13.5% 13.5%2.35%2.35%

The 68­95­99.7 rule

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5.3 and 5.4 ­ Standard Deviation & Normal Distribution (Solutions).notebookDecember 14, 2015

A soft­drink manufacturer has a machine that fills cans.The machine is designed to place an average of 377.0 ml in the can with a standard deviation of 0.5 ml.

34%34%

13.5% 13.5%2.35%2.35%

Example 1

c) Cans containing less than 375.5 ml are rejected.  In a production run of 50 000, how many cans will be rejected?

b) What is the probability that a randomly selected can will contain a volume greater than 376.5 ml?

a) What percent of the cans will have less than:

i) 376.0 mL ii) 375.5 mL

Suppose the tread life of XXT tires is normally distributed with a mean of 72 000 km and a standard deviation of 8000 km. Sketch the normal distribution curve and answer these questions.

a) What percent of the tires have lives between 64 000 km and 80 000 km?

Example 2

b) If the store sold 350 tires,  how many would have a tread life between 56 000 km and 96 000 km?

34%34%

13.5% 13.5%2.35%2.35%

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5.3 and 5.4 ­ Standard Deviation & Normal Distribution (Solutions).notebookDecember 14, 2015

 

• Determine the mean and standard deviation.                                                                           

• Using the mean and the standard deviation and frequency table, determine the percentage of the data that fall with 1 and 2 standard deviations of the mean.

• If these percentages are close to those of a normal curve, the data is normally distributed.

Determining if a set of data is Normally Distributed

Mean = 104.5 minStandard deviation = 22.3 min

FrequencyInterval

59.5 ­ 82.0

82.0 ­ 104.5

104.5 ­ 127.0

127.0 ­ 149.5

Are the lengths of Tiegan's movies normally distributed? Explain 

Example 3: p.       ­  #4 

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5.3 and 5.4 ­ Standard Deviation & Normal Distribution (Solutions).notebookDecember 14, 2015

p. 251

# 1, 3, 7, 8a, 10, 12