Chapter 6: The Standard Deviation as a Ruler and the Normal Model By Marisa Suzuki

Standard Deviation as a Ruler Standard deviations can be used to compare very different-looking data Standard deviation tells us how the whole collection of values varies The more variability in data, the higher the standard deviation will be

Standardizing with z-scores Results can be standardized, resulting in values denoted with the letter z, but are usually called z-scores Z-scores measure the distance of each data value from the mean in standard deviations Can be found using the equation:

Standardizing with z-scores cont. The mean of all of the data is subtracted from the sample value y, which is then divided by the standard deviation. A negative z-score tells us that the data is below the mean, while a positive z-score tells us that the data is above the mean

Combining Two Variables To find the standard deviation of the sum or difference, you must ADD the variances and then take the SQUARE ROOT

Normal Models Distributions that are unimodal and roughly symmetric can be represented by a normal model (bell-shaped curve) N(μ,σ) denotes a normal model µ = (mew) mean of the model σ = (lower case sigma) standard deviation in the model The standard normal model is a model with a mean of 0 and SD of 1, denoted as N(0,1)

Normal Models cont. Normal models give us an idea of how extreme a value is by telling us how likely it is to find one that far from the mean The 68-95-99.7 Rule 68% of the values fall within 1 standard deviation of the mean 95% of the values fall within 2 standard deviations of the mean 99.7% of the values fall within 3 standard deviations of the mean

The Bell-Shaped Curve

Question #35 Based on the normal model for weights of Angus steers N(1152,85), what are the cutoff values for: The highest 10% of the weights? The lowest 20% of the weights? The middle 40% of the weights?

Question #35 (part a) 2ND  DISTR  invNorm (0.9, 0, 1)  1.282 The highest 10% of the weights? 2ND  DISTR  invNorm (0.9, 0, 1)  1.282 z=(y-µ)/σ 1.282=(y-1152)/84 y=1259.7 lbs

Question #35 (part b) 2ND  DISTR  invNorm (0.2, 0, 1)  -0.842 The lowest 20% of the weights? 2ND  DISTR  invNorm (0.2, 0, 1)  -0.842 z=(y-µ)/σ -0.842=(y-1152)/84 y=1081.3 lbs

Question #35 (part c) 2ND  DISTR  invNorm (0.3, 0, 1)  -0.524 The middle 40% of the weights? 2ND  DISTR  invNorm (0.3, 0, 1)  -0.524 2ND  DISTR  invNorm (0.7, 0, 1)  0.524 z=(y-µ)/σ -0.524=(y-1152)/84 y=1108.0 lbs 0.524=(y-1152)/84 y=1196.0 lbs

Question #37 Consider the Angus steer model N(1152,84) again: What weight represents the 40th percentile? What weight represents the 99th percentile? What’s the IQR of the weights of these Angus steers?

Question #37 (part a) z=(y-µ)/σ -0.253=(y-1152)/84 y=1130.7 lbs What weight represents the 40th percentile? 2ND  DISTR  invNorm (0.4, 0, 1)  -0.253 z=(y-µ)/σ -0.253=(y-1152)/84 y=1130.7 lbs

Question #37 (part b) z=(y-µ)/σ 2.326=(y-1152)/84 y=1347.4 lbs What weight represents the 99th percentile? 2ND  DISTR  invNorm (0.99, 0, 1)  2.326 z=(y-µ)/σ 2.326=(y-1152)/84 y=1347.4 lbs

Question #37 (part c) What’s the IQR of the weights of these Angus steers? 2ND  DISTR  invNorm (0.25 0, 1)  -0.674 2ND  DISTR  invNorm (0.75, 0, 1)  0.674 Q1 = 1095.34 Q3 = 1208.60 IQR = (Q3 – Q1) = (1208.60 – 1095.34) = 113.3 lbs