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Chapter 4 The Normal Distribution EPS 625 Statistical Methods Applied to Education I

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Chapter 4 The Normal Distribution EPS 625 Statistical Methods Applied to Education I. Figure 4.3, page 85. Beginning on Page 634. What is the proportion of scores in a normal distribution between the mean and z = +0.52?. Answer: 19.85%. What is the proportion of scores

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Presentation Transcript
slide1

Chapter 4

The Normal Distribution

EPS 625

Statistical Methods Applied to Education I

slide8

What is the proportion of scores

in a normal distribution

between the mean and z = +0.52?

Answer: 19.85%

slide9

What is the proportion of scores

in a normal distribution

between the mean and z = -1.89?

Answer: 47.06%

slide10

What is the proportion of scores

in a normal distribution

between z = -0.19 and z = +3.02?

Area A = 7.53% (and) Area B = 49.87%

Therefore, the Area of Interest = 7.53 + 49.87 = 57.40%

slide11

What is the proportion of scores

in a normal distribution

between z = +0.19 and z = +1.12?

Area A = 7.53% (and) Area B (total) = 36.86%

Therefore, the Area of Interest* = 36.86 - 7.53 = 29.33%

slide13

What is the proportion of scores

in a normal distribution

between z = -1.09 and z = -3.02?

Area A = 36.27% (and) Area B (total) = 49.87%

Therefore, the Area of Interest* = 49.87 – 36.27 = 13.60%

slide14

What is the proportion of scores

in a normal distribution above z = +0.87?

Area A (total beyond z = 0.00) = 50.00% (and) Area B = 30.78%

Therefore, the Area of Interest = 50.00 – 30.78 = 19.22%

Or – use “Area Beyond z” Column

Locate z = 0.87 and you will find 19.22%

slide15

What is the proportion of scores

in a normal distribution below z = +1.28?

We know that Area A (total beyond z = 0.00) = 50.00%

We find Area B = 39.97%

Therefore, the Area of Interest = 50.00 + 39.97 = 89.97%

slide16

What is the proportion of scores

in a normal distribution above z = -2.00?

We know that Area B (total beyond z = 0.00) = 50.00%

We find Area A = 47.72%

Therefore, the Area of Interest = 50.00 + 47.72 = 97.72%

slide17

What is the proportion of scores

in a normal distribution below z = -0.52?

Area A = 19.85% (and) Area B (total beyond z = 0.00) = 50.00%

Therefore, the Area of Interest = 50.00 – 19.85 = 30.15%

Or – use “Area Beyond z” Column

Locate z = 0.52 and you will find 30.15%