Bellwork Suppose that the average blood pressures of patients in a hospital follow a normal distribution with a mean of 108 and a standard deviation of 14. Set up a normal distribution curve. What percentage of patients have a blood pressure of 94 or higher? What percentage of patients have a blood pressure between 80 and 122? What percentage of patients have a blood pressure of less that 80?

Z-Scores

Standard Score or “Z-Score” The “Z-score” tells how many standard deviations from the mean an observation falls. So far we have only discussed standard deviations of 1, 2, and 3 away from the mean. What if it is 1.24 or -3.15 away from the mean. Shouldn’t we be able to calculate the population percentages with different deviations?

Z-Score To calculate the z-score, use the following: 𝑧 = z-score 𝑥 = data point you are interested in 𝑥 =  = mean (use full number from calculator) Sx =  = Standard Deviation (use full number from calculator) 𝑧 = 𝑥− 𝑥 𝑆 𝑥 = 𝑥−𝜇 𝜎

Using the following set of data, find the z-scores for 52, 89, and 25: 59, 27, 18, 78, 61, 91, 52, 34, 54, 93, 100, 87, 85, 82, 68 Mean = 65.93333333 Standard Deviation = 25.32493595

𝑧= 52−65.93333333 25.32493595 = -.5501823719 What does this mean?? Well, it means that 52 is a little over one-half of a standard deviation below the mean. This should make sense. The standard deviation is about 25 and the mean is about 66. The difference between 66 and 52 is 14 which is about half of 25.

Well, it means that 89 is almost 1 standard deviation above the mean. 𝑧= 89−65.93333333 25.32493595 = .9108282333 What does this mean?? Well, it means that 89 is almost 1 standard deviation above the mean. This should make sense. The standard deviation is about 25 and the mean is about 66. The difference between 66 and 89 is 23 which is almost 25.

𝑧= 25−65.93333333 25.32493595 = -1.616325246 What does this mean?? Well, it means that 25 is a little over one and one-half standard deviations below the mean. This should make sense. The standard deviation is about 25 and the mean is about 66. The difference between 66 and 25 is 41 which is about 1.5*25 = 37.5.

Comparing Different Data The entire reason to go through all of this is to be able to compare two different types of data and put them on level playing fields. Comparing how a student did on a history test to how a student did on a math test are certainly not comparable at face value. However, if we have means and standard deviations, we can start comparing these two seemingly different things.

Example 1 The single-season home run record for Major League Baseball has been set just three times since Babe Ruth hit 60 home runs in 1927. Roger Maris hit 61 in 1951, Mark McGwire hit 70 in 1998, and Barry Bonds hit 73 in 2001. In an absolute sense, Barry Bonds had the best performance of these four players because he hit the most home runs in a single season. However, in a relative sense, this may not be true. GRAY AREA!!

Baseball historians suggest that hitting a home run has been easier in some eras than others. This is due to many factors, including the quality of batters, quality of pitchers, hardness of the baseball, dimensions of ballparks, and possible use of performance-enhancing drugs. To make a fair comparison, we should see how these performances rate relative to those of other hitters during the same year.

Compute the standardized scores for each performance using the information in the table. Year Player HR Mean SD Z-Score 1927 Ruth 60 7.2 9.7 1961 Maris 61 18.8 13.4 1998 McGwire 70 20.7 12.7 2001 Bonds 73 21.4 13.2 𝑧= 60−7.2 9.7 =5.44 𝑧= 61−18.8 13.4 =3.15 𝑧= 70−20.7 12.7 =3.88 𝑧= 73−21.4 13.2 =3.91 Although all four performances were outstanding, Babe Ruth can still lay claim to being the single-season home run champ! … and, NO, Mrs. Rivera did not witness him playing -- she was not alive back then!

So the probability of falling with 1 So the probability of falling with 1.43 standard deviations from the mean of the heights is 84.72%

Look at table and find closest to .98  2.05 2.05= 𝑥−100 16  x – 100 = 2.05(16)  x = 2.05(16) + 100  x = 132.8

650 on the SAT (mean = 500, Std. Dev. = 100) 30 on the ACT (mean = 21, Std. Dev. = 4.7) 𝑧= 650−500 100 =1.5 1.5 on the table is .9332 = 93.32% or 93rd percentile 𝑧= 30−21 4.7 =1.91 1.91 on the table is .9719 = 97.19% or 97th percentile Conclusion: A 30 on the ACT is better than a 650 on the SAT due to percentile ranking being higher!

42.34% 69.49 inches

Normalcdf(-10000000, 650, 500, 100) Normalcdf(-10000000, 30, 21, 4.7) 650 on the SAT (mean = 500, Std. Dev. = 100) 30 on the ACT (mean = 21, Std. Dev. = 4.7) Normalcdf(-10000000, 650, 500, 100) .9331927713 = 93.32% or 93rd percentile Normalcdf(-10000000, 30, 21, 4.7) .9722470321 = 97.22% or 97th percentile Conclusion: normalcdf is more accurate than the tables but the tables are very close!

invNorm(98, 100, 16) =132.86

Example 1 The duration of routine operations in a certain hospital has approximately a normal distribution with an average of 125 minutes and a standard deviation of 18 minutes. Set up a normal distribution curve. a) What percentage of operations last at least 125 minutes? b) What percentage of operations lasted between 107 and 143 minutes? c) What percentage of operations lasted between 89 and 125 minutes? Normalcdf(-10000000, 125, 125, 18) = .50 Normalcdf(107, 143, 125, 18) = .68268948 Normalcdf(89, 125, 125, 18) = .477249938

Example 2 The mean weight of 850 college students is 70 kg and the standard deviation of 3 kg. Set up a normal distribution curve. Determine how many students weigh: Between 64 kg and 73 kg. More than 76 kg. Less than 67 kg. Between 70 kg and 76 kg. Normalcdf(64, 73, 70, 3) *850 = 695 Normalcdf(76, 10000000, 70, 3) *850 = 19 Normalcdf(-10000000, 67, 70, 3) *850 = 134 Normalcdf(70, 76, 70, 3) *850 = 405