Click on the graph in the upper right and drag the cursor to move the shaded

Click on the graph in the upper right and drag the

This preview shows page 60 - 66 out of 96 pages.

Click on the graph in the upper right and drag the cursor to move the shaded area on the graph. Move the cursor until the shaded region represents the probability of interest (the “Between (Red-Shaded)” number shown in the Probabilities column on the right side will show the probability). The value on the horizontal axis that marks the right side of the shaded area will be the percentile of interest. 49/77
Finding Percentiles Using The Normal Table Note: Confusingly, some people refer to the q from the q th percentile as the percentile, rather than the value that corresponds to that probability. Finding percentiles from a Normal table To find the q th percentile of a N ( μ, σ ) distribution: 1 Use the table for the N (0 , 1) distribution to find the z -score of the q th percentile. 2 Convert the z -score to proper units. Since the z -score is defined as z = value - mean standard deviation = x - μ σ , then solving for x means the q th percentile is q th percentile = x = μ + σz. 50/77
Example: Children’s Heights III Example Suppose children’s heights follow a N (68 , 2) distribution. How tall does a child need to be in order to be in the 90th percentile? z .05 .06 .07 .08 .09 0.8 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8289 0.8315 0.8340 0.8365 0.8389 1 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9265 0.9279 0.9292 0.9306 0.9319 51/77
Example: Children’s Heights III Example Suppose children’s heights follow a N (68 , 2) distribution. How tall does a child need to be in order to be in the 90th percentile? z .05 .06 .07 .08 .09 0.8 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8289 0.8315 0.8340 0.8365 0.8389 1 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9265 0.9279 0.9292 0.9306 0.9319 To find the 90th percentile, we want to find the z -score that corresponds to a probability of 0.90. From the Normal table, the probability of 0.8997 is the closest to 0.90. The probability of 0.8997 corresponds to a z -score of 1.28. 51/77
Example: Children’s Heights III Example (Continued) Suppose children’s heights follow a N (68 , 2) distribution. How tall does a child need to be in order to be in the 90th percentile? To find the 90th percentile, we want to find the z -score that corresponds to a probability of 0.90. From the Normal table, the probability of 0.8997 is the closest to 0.90. The probability of 0.8997 corresponds to a z -score of 1.28. Converting the z -score of 1.28 into inches, we have z = 1 . 28 = x - μ σ = x - 68 2 , so x = μ + σz = 68 + (2 × 1 . 28) = 70 . 56 . A child needs to be 70.56 inches tall to be in the 90th percentile of heights. 52/77
The Normal Model and the Empirical Rule Recall from Chapter 3: The Empirical Rule If the distribution is symmetric and unimodal, then Approximately 68% of the observations (roughly two-thirds) will be within one standard deviation of the mean. Approximately 95% of the observations will be within two standard deviations of the mean. Nearly all the observations will be within three standard deviations of the mean. The Empirical Rule applies for any arbitrary symmetric and unimodal distribution as an approximate guideline. However, for the Normal model, the Empirical Rule is (nearly) exact.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture