Date: 07/05/97 at 05:49:21 From: Jeremy Bowell Subject: Averages involving standard deviation How do I solve this problem? The average height of year 10 students is given as 175 cm and the standard deviation is 12cm. Find the percentage of students whose height is: a) Greater than 175 cm b) Between 163 and 187 cm c) Greater than 187 cm d) Between 151 and 163 cm What is standard deviation?
Date: 07/05/97 at 07:58:37 From: Doctor Anthony Subject: Re: Averages involving standard deviation Standard deviation is a measure of the spread of a distribution about the mean. It is the square root of the VARIANCE, where variance is calculated from: VAR(x) = E(x^2) - Mean^2 E(x^2) is the 'expected value' of x^2. If this term is unfamiliar to you, consult a standard textbook or write back. (a) Because the normal distribution is symmetric about the mean, the percentage of students with height greater than the mean, 175 cm, is 50 percent. Using the normal tables, let m = mean and s = standard deviation. Z values are those on the horizontal axis giving the number of standard deviations from the mean, with z = 0 at the mean. Areas are the areas under the normal probability curve between two z values. These areas are found using normal tables and entering the appropriate z values. (b) You calculate the z values using: x - m 163 - 175 z = ------ = --------- = -1 s 12 The area between the mean and -1 s.d. = .3413 187 - 175 Also for the upper limit z = ----------- = +1 12 Again, the area between the mean and 1 s.d. = .3413 Total probability is then 2 x .3413 = 0.6826, which is 68.26 percent. (c) Greater than 187 gives area 0.5 - .3413 = .1587, so the number greater than 187 is 15.87 percent. (d) 151 - 175 z = --------- = -2 area between mean and -2 is 0.47725 12 163 - 175 z = ---------- = -1 area between mean and -1 is 0.3413 12 The area between -2 and -1 is then .47725 - .3413 = 0.13595 So 13.6 percent of students have heights between 151 and 163 cms. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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