STAT1008 Lecture Notes - Lecture 22: Standard Score, Standard Deviation, Percentile
STAT1008 Week 8 Lecture A
● How do we find areas under a normal density?
○ Know the mean and standard deviation of the normal distribution
○ Can use calculus (finding area under the curve)
○ E.g. Verbal SAT scores
■ Suppose that verbal SAT scores for applicants at a college follow a
normal distribution with mean mu = 580 and stn. Dev = 70
■ What proportion of applicants have SAT score above 700?
■ R: pnorm (700, mean=580, sd=70, lower.tail=FALSE) = 0.043
○ E.g. Bootstrap means
■ Suppose that the bootstrap distribution of means for samples size 500
Atlanta commute times is N(29.11, 0.93)
■ Find an endpoint (percentile) so that just 5% of the bootstrap means are
smaller thus from R: qnorm(0.05, mean=29.11, sd=0.93) = 27.58029
● Finding probabilities for N(mu, std dev):
○ About what proportion should be within one std dev of the mean?
○ Note: All that really matters is the number of standard deviations from the mean
● Standard Normal:
○ Mean = 0, Standard deviation = 1 thus Z~N(0,1)
○ To convert any X~N(mu,standard deviation) to Z~N(X-mu, standard deviation) <-
Z score
○ Z score = standard normal
○ Standardise the endpoints, then use Z~N(0,1)
○ E.g. Dog/Owner randomisation proportions, p hat approx N(0.5,0.1)
■ X-area above 0.64 = Z-area above 0.64-0.5/0.1 = 1.40
■ 1.40 represents the area on the LHS of 0.64
● Converting Normals:
○ X~N(mean, standard dev) -> Z= X-mu/standard dev -> Z~N(0,1) -> X = hu + Z x
Std Dev -> X~N(mean, standard dev)
○ E.g. Percentile for verbal SAT
■ 25% percentile (Q1) for Z~N(0,1) is -0.674
■ Find the 25% percentile for the Verbal SAT~N(580,70) distribution
● X = mu + Z x stn dev = 580 + (-0.674)(70) = 533
● Standardings:
○ If X~N(mu, stn dev) then use the linear transformation Z= X-mu/stn dev ~ N(0,1)
○ The resulting standardised value is sometimes called the “z value” or “z score”
○ For Z~N(0,1), values for Pr(Z< or equal to z) are available in tables
○ So, for ANY random variable that comes from a normal distribution, if we subtract
the mean and divide by the stn dev, we get a r.v.~N(0,1)
○ Area in tables - Pr(Z<z) = area to the left of z
find more resources at oneclass.com
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Document Summary
Know the mean and standard deviation of the normal distribution. Can use calculus (finding area under the curve) Suppose that verbal sat scores for applicants at a college follow a normal distribution with mean mu = 580 and stn. R: pnorm (700, mean=580, sd=70, lower. tail=false) = 0. 043. Suppose that the bootstrap distribution of means for samples size 500. Find an endpoint (percentile) so that just 5% of the bootstrap means are smaller thus from r: qnorm(0. 05, mean=29. 11, sd=0. 93) = 27. 58029. Note: all that really matters is the number of standard deviations from the mean. Mean = 0, standard deviation = 1 thus z~n(0,1) To convert any x~n(mu,standard deviation) to z~n(x-mu, standard deviation) z= x-mu/standard dev -> z~n(0,1) -> x = hu + z x. 25% percentile (q1) for z~n(0,1) is -0. 674.
