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Normal distribution percentages worksheet10/21/2023 ![]() Therefore, the weight that includes 85% and below of the total population is approximately 166.6 lb. The study revealed that the spending distribution is approximately normally distributed with a mean of 4.11 and a standard deviation of 1.37. Exercise 11.3.1 If the area to the left of x is 0.012, then what is the area to the right Answer Example 11.3.2 The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five. ![]() Weight = mean + (z-score * standard deviation) Calculate Example 11.3.1 If the area to the left is 0.0228, then the area to the right is 1 0.0228 0.9772. The formula to find the weight corresponding to a given z-score is: Using a standard normal distribution table, we find that the z-score corresponding to the 85th percentile is approximately 1.04. (c) the BMI values that correspond to the middle 99.7 of the. (b) the percentages of males with BMI greater than 12.3. Use the 68-95-99.7 rule to nd (a) the percentage of males with BMI less than 20.1. The solutions to these problems are at the bottom of the page. The BMI for males age 20 to 74 is follows approximately a normal distribution with mean 27.9 and standard deviation 7.8. Problems and applications on normal distributions are presented. To find the weight that includes 85% and below of the total population, we need to find the z-score corresponding to the 85th percentile of the normal distribution and then use it to find the corresponding weight. Normal Distribution Problems with Solutions. A specific percentage of the data is assigned to each standard deviation away from the mean on a normal distribution. Therefore, the probability that a randomly selected male student weighs less than 128 lb is 0.0630 or approximately 6.3%.Ĭ. Using a standard normal distribution table, we find that the area to the left of z = -1.53 is 0.0630. To find the probability that a randomly selected male student weighs less than 128 pounds, we need to calculate the z-score for a weight of 128 lb and then use a standard normal distribution table to find the area to the left of that z-score. Therefore, approximately 0.6443 * 500 = 322 students weigh between 120 and 155 lb.ī. Step 2: Divide the difference by the standard deviation. The standard normal distribution is a bell-shaped planar region bounded above by a bell-shaped curve and below by a horizontal line representing the z-values of the normal distribution. ![]() Step 1: Subtract the mean from the x value. A key tool for calculating proportions and percentages associated with normal distri-butions is the standard normal distribution. 2 83, point, 2 points and a standard deviation of 8 8 8 8 points. The z score tells you how many standard deviations away 1380 is from the mean. A set of history exam scores are normally distributed with a mean of 83.2 83.2 8 3. Using a standard normal distribution table, we find that the area between z = -2.07 and z = 0.27 is 0.6443. Normal distribution: Area between two points. To find the number of students who weigh between 120 and 155 lb, we need to calculate the z-scores for each weight and then use a standard normal distribution table to find the area between the two z-scores.
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