Working with z-scores and the Normal Curve... it's everywhere!

The Bell Curve by flickr user c r i s

... but first a little quiz ... ready?

Studying by scui3asteveo

The table shows the lengths in millimetres of 52 arrowheads.16 16 17 17 18 18 18 18 19 20 20 21 2121 22 22 22 23 23 23 24 24 25 25 25 2626 26 26 27 27 27 27 27 28 28 28 28 2930 30 30 30 30 30 31 33 33 34 35 39 40

(a) Calculate the mean length and the standard deviation.

(d) What percent of the arrowheads are within one standard deviation of the mean length?

(c) How many arrowheads are within one standard deviation of the mean?

(b) Determine the lengths of arrowheads one standard deviation below and one standard deviation above the mean.

HOMEWORK

North American women have a mean height of 161.5 cm and a standard deviation of 6.3 cm.

(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?

(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

179.14 = x

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 high must she set the passing mark if she wants only the top 75 percent of the marks to be passing marks?

• how high must she set the mark for an 'A' if she wants 5 percent of the students to get an A?

• how many students should have failed (i.e., less than 50 percent)?

Curving The Marks

Working with the Normal DistributionExample: 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%

Working with the Normal DistributionExample: 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)

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.

The Standard Normal Distribution

ShadeNorm(Lo z, Hi z)[shades area under std. normal curve]

ShadeNorm(Lo value, Hi value, mean, std. dev.)[shades area under modified normal curve]

µ+5σ

µ-5σ

σ

σ

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.

Case 1(a): Calculate the Percentage of Scores Between Two Given Scores HOMEWORK

Find the percent of z-scores in a standard normal distribution that are:

(d) between z = 0.55 and z = 0.15(c) between z = -1.11 and z = 0.92

(b) above z = -2.35(a) below z = 0.52

HOMEWORK

Find the z-score if the area under a standard normal curve:

(c)to the right of z is 0.785

(b) to the right of z is 0.305

(a) to the left of z is 0.812

HOMEWORK

Determine the values for z and x. HOMEWORK