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Math 20-2 Statistics: Lesson #6 Z-Scores Objective: By the end of the lesson, you should be able to: - Determine the z-score of a given value in a normally distributed data set. - Explain the meaning of a z-score. Key Point: Z-scores allow us to compare data from different normal distributions. We have seen that different standard deviations and different means can change the shape and placement of a normal distribution curve. Sometimes we need to compare data from two different normal distributions. We can do this by finding out how many standard deviations the data values are from the mean. This is called the _____________ of the data value. e.g. 1) a) A z-score of 2.3 means the data value is 2.3 standard deviations ______________ the mean. b) A z-score of –1.9 means the data value is 1.9 standard deviations ______________ the mean. The formula for finding the z-score of a given data value is: z Where: = = x = (*This formula is on your formula sheet.) e.g. 2) A set of data has a mean of 15 and a standard deviation of 2.4. Find the z-score for the following data values. (Round to the nearest hundredth.) a) 12 Math 20-2 Statistics: Lesson #6 b) 20.5 c) 15 You can also use the z-score formula to find the mean ( ), standard deviation ( ), or data value (x) if you are given the other values. e.g. 3) The height of men between the ages of 18 and 24 is normally distributed with a mean of 180 cm and a standard deviation of 8 cm. Todd’s height has a z-score of -2.2. How tall is he? e.g. 4) The marks on a diploma exam for Biology 30 are normally distributed. Susan scored 68%, which was 0.35 standard deviations above the mean. If the mean of the exam is 64%, then find the standard deviation to the nearest tenth of a percent. Math 20-2 Statistics: Lesson #6 Remember: Z-scores are helpful in comparing data from different distributions. e.g. 5) Hailey belongs to a running club in Vancouver. The members of the club took a trip to Lake Louise, where they ran a 200 m sprint. The run times for the club were normally distributed. The table below shows Hailey’s run times in Vancouver and Lake Louise, as well as the club mean and standard deviation: Club Mean Club Standard Hailey’s Run Location Time (s) Deviation (s) Time (s) Vancouver 25.75 0.62 24.95 Lake Louise 25.57 0.60 24.77 How could you determine in which location Hailey’s run time was better, compared to the club results? Explain. e.g. 6) Paul got a mark of 75% on a math test with a mean of 61% and a standard deviation of 12%. He got 72% on a chemistry exam with mean 63% and standard deviation 7%. If the marks on both tests were normally distributed, on which test did he do better relative to the class? Assignment: p. 292-293 #5, 10, 12