Normal distributions Normal curves are used to model many biological variables. They can describe a population distribution or a probability distribution. Normal—or Gaussian—distributions are a family of symmetrical, bell- shaped density curves defined by a mean  (mu) and a standard deviation  (sigma): N(  ). xx Inflection point

A family of density curves Here means are different (  = 10, 15, and 20) whereas standard deviations are the same (  = 3). Here means are the same (  = 15) whereas standard deviations are different (  = 2, 4, and 6).

Human heights, by gender, can be modeled quite accurately by a Normal distribution.   Guinea pigs survival times after inoculation of a pathogen are clearly not a good candidate for a Normal model!

 About 68% of all observations are within 1 standard deviation (  of the mean (  ).  About 95% of all observations are within 2  of the mean .  Almost all (99.7%) observations are within 3  of the mean. The 68–95–99.7 rule for any N(μ,σ) Number of times σ from the center µ All normal curves N(µ,σ) share the same properties: To obtain any other area under a Normal curve, use either technology or Table B.

Population of young adults N(0,1) Standardized bone density (no units) What percent of young adults have osteoporosis or osteopenia? World Health Organization definitions of osteoporosis based on standardized bone density levels Normal Bone density is within 1 standard deviation (z > –1) of the young adult mean or above. Low bone mass (osteopenia) Bone density is 1 to 2.5 standard deviations below the young adult mean (z between –2.5 and –1). Osteoporosis Bone density is 2.5 standard deviations or more below the young adult mean (z ≤ –2.5). These are bone densities of  1 or less, representing the area to the left of the middle 68% between  1 and +1. So ???%.

Standardized bone density (no units) Women aged 70 to 79 are NOT young adults. The mean bone density in this age is about −2 on the standard scale for young adults. What is the probability that a randomly chosen woman in her 70s has osteoporosis or osteopenia (< −1)??? A) 97.5% B) 95% C) 84% D) 68% E) 50% Young adults N(0,1)Women N(-2,1)

We can standardize data by computing a z-score: If x has the N(  ) distribution, then z has the N(0,1) distribution. N(0,1) => N(64.5, 2.5) Standardized height (no units) The standard Normal distribution

A z-score measures the number of standard deviations that a data value x is from the mean . Standardizing: z-scores When x is larger than the mean, z is positive. When x is smaller than the mean, z is negative. When x is 1 standard deviation larger than the mean, then z = 1. When x is 2 standard deviations larger than the mean, then z = 2.

mean µ = 64.5" standard deviation  = 2.5" height x = 67" We calculate z, the standardized value of x: Given the rule, the percent of women shorter than 67” should be, approximately, ???%. Area= ??? N(µ,  ) = N(64.5, 2.5)  = 64.5” x = 67” z = 0z = 1 Women’s heights follow the N(64.5”,2.5”) distribution. What percent of women are shorter than 67 inches tall (that’s 5’6”)?

Using Table B (…) Table B gives the area under the standard Normal curve to the left of any z-value is the area under N(0,1) left of z = – is the area under N(0,1) left of z = – is the area under N(0,1) left of z = –2.56

Area ≈ 0.84 Area ≈ 0.16 N(µ,  ) = N(64.5, 2.5)  = 64.5 x = 67 z = 1  ???% of women are shorter than 67”.  Therefore, ???% of women are taller than 67" (5'6"). For z = 1.00, the area under the curve to the left of z is

Tips on using Table B Because of the curve’s symmetry, there are two ways of finding the area under N(0,1) curve to the right of a z-value. area right of z = 1 – area left of z Area = Area = z = area right of z = area left of –z

Using Table B to find a middle area To calculate the area between two z-values, first get the area under N(0,1) to the left for each z-value from Table B. area between z 1 and z 2 = area left of z 1 – area left of z 2 Don’t subtract the z-values!!! Normal curves are not square! Then subtract the smaller area from the larger area.  The area under N(0,1) for a single value of z is zero.

The blood cholesterol levels of men aged 55 to 64 are approximately Normal with mean 222 mg/dl and standard deviation 37 mg/dl. What percent of middle-age men have high cholesterol (> 240 mg/dl)??? What percent have elevated cholesterol (between 200 and 240 mg/dl)???? xz area left area right %31% %72%

Inverse Normal calculations You may also seek the range of values that correspond to a given proportion/ area under the curve. For that, use technology or use Table B backward.  First find the desired area/ proportion in the body of the table,  then read the corresponding z-value from the left column and top row. For a left area of 1.25% (0.0125), the z-value is –2.24.

The hatching weights of commercial chickens can be modeled accurately using a Normal distribution with mean μ = 45 g and standard deviation σ = 4 g. What is the third quartile of the distribution of hatching weights? Q 3 ≈ 47.7 g We know μ, σ, and the area under the curve; we want x. Table B gives the area left of z  look for the lower 25% We find z ≈ 0.67

The blood cholesterol levels of men aged 55 to 64 are approximately normal with mean 222 mg/dl and standard deviation 37 mg/dl. What range of values corresponds to the 10% highest cholesterol levels??? Cholesterol level (mg/dl) A) > 175 B) > 247 C) > 269 D) > 288 z Area left Area right %10%

One way to assess if a data set has an approximately Normal distribution is to plot the data on a Normal quantile plot. The data points are ranked and the percentile ranks are converted to z- scores. The z-scores are then used for the horizontal axis and the actual data values are used for the vertical axis. Use technology to obtain Normal quantile plots. If the data have approximately a Normal distribution, the Normal quantile plot will have roughly a straight-line pattern. Normal quantile plots

Roughly normal (~ straight-line pattern) Right skewed (most of the data points are short survival times, while a few are longer survival times)