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# Applied 40S April 7, 2009

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Introduction to the 3 different "meanings" of "area under the normal curve" and the ShadeNorm() command on the TI graphing calculators.

Introduction to the 3 different "meanings" of "area under the normal curve" and the ShadeNorm() command on the TI graphing calculators.

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• 1. Working with z-scores and the Normal Curve ... it's everywhere! The Bell Curve by ﬂickr user c r i s
• 2. ... but ﬁrst a little quiz ... ready? Studying by scui3asteveo
• 3. HOMEWORK The table shows the lengths in millimetres of 52 arrowheads. 16 16 17 17 18 18 18 18 19 20 20 21 21 21 22 22 22 23 23 23 24 24 25 25 25 26 26 26 26 27 27 27 27 27 28 28 28 28 29 30 30 30 30 30 30 31 33 33 34 35 39 40 (b) Determine the lengths of arrowheads (a) Calculate the mean length one standard deviation below and one and the standard deviation. standard deviation above the mean. (d) What percent of the arrowheads (c) How many arrowheads are within one standard deviation are within one standard of the mean length? deviation of the mean?
• 4. North American women have a mean height of 161.5 cm and a standard deviation of 6.3 cm. (a) A car designer designs car seats to ﬁt women taller than 159.0 cm. What is the z-score of a woman who is 159.0 cm tall? z = -0.3968 (b) The manufacturer designs the seats to ﬁt women with a maximum z-score of 2.8. How tall is a woman with a z-score of 2.8? 179.14 = x
• 5. Curving The Marks Professor Adams has 140 students who wrote a statistics test. If the marks are approximately normally distributed: • how many students should have a 'B' mark (i.e., 70 to 79 percent)? • how many students should have failed (i.e., less than 50 percent)? • how high must she set the mark for an 'A' if she wants 5 percent of the students to get an A? • how high must she set the passing mark if she wants only the top 75 percent of the marks to be passing marks?
• 6. Working with the Normal Distribution Example: The graph at right represents the marks of a large number of students where the mean mark is 69.3 percent and the standard deviation is 7 percent, and the distribution of marks is approximately normal. 62.3% 69.3% 76.3%
• 7. Working with the Normal Distribution Example: The graph at right represents the marks of a large number of students where the mean mark is 69.3 percent and the standard deviation is 7 percent, and the distribution of marks is approximately normal. We know that if the marks are approximately normally distributed, then approximately: • 68 percent of the marks are between 62.3 percent and 76.3 percent (i.e., µ ± 1σ) • 34% of the marks are between 69.3 percent and 76.3 percent (i.e., between µ and µ + 1σ) • 50% of the marks are below 69.3 percent (i.e., less than µ) • 16% of the marks are above 76.3 percent (i.e., greater than µ + 1σ)
• 8. The Standard Normal Distribution The normal distributions used up to this point have all been non- standard normal distributions. This means that the mean and standard deviation of the distribution are the mean and standard deviation of the data being studied. In a standard normal distribution, the scale on the x-axis is the z- score (a standard score) where 'z = 0' is the mean, and the standard deviation is '1'. The distribution is a probability distribution where the area under the curve = 1. This means there is a 100 percent chance of every score being included in this distribution. Other probabilities are illustrated below.
• 9. ShadeNorm(Lo z, Hi z) [shades area under std. normal curve] σ ShadeNorm(Lo value, Hi value, mean, std. dev.) [shades area under modiﬁed normal curve] µ-5σ σ µ+5σ
• 10. Case 1(a): Calculate the Percentage of Scores Between Two Given Scores HOMEWORK The mean mark for a large number of students is 69.3 percent with a standard deviation of 7 percent. What percent of the students have a 'B' mark (i.e., 70 percent to 79 percent)? Assume that the marks are normally distributed.
• 11. HOMEWORK Find the percent of z-scores in a standard normal distribution that are: (a) below z = 0.52 (b) above z = -2.35 (d) between z = 0.55 and z = 0.15 (c) between z = -1.11 and z = 0.92
• 12. HOMEWORK Find the z-score if the area under a standard normal curve: (a) to the left of z is 0.812 (b) to the right of z is 0.305 (c)to the right of z is 0.785
• 13. HOMEWORK Determine the values for z and x.