 Show Since the tail area for the lowest 25% has a z= -0.67 and the z for the upper 25% is z = 0.67, the interquartile range is 1.34. 1.5*1.34=2.01. So we need to find the percent that falls above .67+2.01 = 2.68 and the percent that falls below -2.68. The tail area is 2*0.0037 = 0.0074 = .74%. Note: In this problem, we based our calculations for the IQR on the population distribution (the normal curve). When making a boxplot, the IQR is based on the sample. The sample has variability in it, also, and so in practice we identify more than .74% of our observations as outliers. 2. Imagine two outstanding students who have excelled in academics, service, and athletics. Anita earned 620 on the SAT verbal and Bonita scored 27 on the ACT. In other respects, their high school records are comparable. You learn that the average SAT verbal score is 500, with standard deviation of 100 points while the ACT has a mean of 18 with a standard deviation of 6. If we assume that scores on these exams are approximately normal, can you use thier scores to determine who wins the scholarship? Bonita wins since her score is 1.5SD's above average and Anita's score is 1.2 SD's above average. 3. Anita scored at the 60th percentile on a midterm exam while Bonita scores at the 85th percentile. If we know that the midterm had a mean of 70 and standard deviation of 10 points, how many points separate Anita and Bonita on this exam? Anita's score =
70+.26(10) = 72.6 4. National Fruit Company claims that the weights of their forty pound boxes of imported bananas are approximately normal with mean 41 pounds and standard deviation of 4 ounces. How often will a box of bananas be underweight? 5. For women athletes at
UCLA, 62.5 inches is the 25th percentile height and 65.5 inches is the 75th percentile height. Using the equations 65.6 = mu + 0.67sigma and 62.5 = mu - 0.67 sigma, we have 128 = 2mu or mu = 64 inches and sigma = 2.24 inches. Then the 90th percentile height is 64+1.28(2.24) = 66.87 inches. You can use the following formula to calculate the percentile of a normal distribution based on a mean and standard deviation: Percentile Value = μ + zσ where:
The following examples show how to use this formula in practice. Example 1: Calculate 15th Percentile Using Mean & Standard DeviationSuppose the weight of a certain species of otters is normally distributed with a mean of μ = 60 pounds and standard deviation of σ = 12 pounds. What is the weight of an otter at the 15th percentile? To answer this, we must find the z-score that is closest to the value 0.15 in the z table. This value turns out to be -1.04: We can then plug this value into the percentile formula:
An otter at the 15th percentile weighs about 47.52 pounds. Note: We could also use the Percentile to Z-Score Calculator to find that the exact z-score that corresponds to the 15th percentile is -1.0364. Pugging this value into the percentile formula, we get:
Example 2: Calculate 93rd Percentile Using Mean & Standard DeviationSuppose the exam scores on a certain test are normally distributed with a mean of μ = 85 and standard deviation of σ = 5. What is the exam score of a student who scores at the 93rd percentile? To answer this, we must find the z-score that is closest to the value 0.93 in the z table. This value turns out to be 1.48: We can then plug this value into the percentile formula:
A student who scores at the 93rd percentile would receive an exam score of about 92.4. Note: We could also use the Percentile to Z-Score Calculator to find that the exact z-score that corresponds to the 93rd percentile is 1.4758. Pugging this value into the percentile formula, we get:
Additional ResourcesHow to use the Z Table (With Examples) What is the zComputing Percentiles. What percentage of a normal distribution is located between z =The percentage of the normal distribution look at it between -0.45 and -1.32 is 23.30% rate. So that will be the answer for the given question. It just helps you. Thank you for watching.
What is the mean for the following sample of scores scores 1 4 5 6?Sum of scores divided by sample size is called mean. Mean is a central tendency measure. where, ¯x is mean, Σxi Σ x i is the sum of scores, n is the sample size. The mean of given data is 76.
Which zThus, a score that is located two standard deviations above the mean will have a z-score of +2.00. And, a z-score of +2.00 always indicates a location above the mean by two standard deviations.
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A distribution is normal and has μ = 90 and σ = 10. what is the 64th percentile?
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