Statistics (Essay Sample)

A simple random sample of five months of sales data provided the following information Month 1 2 3 4 5 Units sold 94 100 85 94 92 Develop a point estimate of the population mean number of units sold per month. Mean = units soldnumber of months = {94+100+85+94+92}5 = 4655 = 93 Develop a point estimate of the population standard deviation. Let units sold be represented by x x (x-mean) (x-mean)2 94 1 1 100 7 49 85 -8 64 94 1 1 92 -2 4 Standard deviation = (x-mean)2number of months-1 = (1+49+64+1+4)5-1 = 1164 = 29 = 5.3852 The college board reported the following mean scores for three parts of the scholastic aptitude test (SAT) (The World Almanac, 2009): Critical reading 502 Mathematics 515 Writing 494 Assume that the population standard deviation on each part of the test is 100 What is the probability a random sample of 90 test takers will provide a sample mean test score within 10 points of population means of 502 on the critical reading part of the test? It ranges in ±10 502 + 10 = 492 502 – 10 = 512 492≤502≤512 P(492≤502≤512 ) = P(502≤512)- P(502≤492) = P( Z ≤1010090) – P(Z ≤-1010090) = P(Z≤0.95)- P(Z≤-0.95) = 0.8289 – 0.1711 = 0.6578 Therefore P(492≤502≤512 ) = 0.6578 What is the probability a random sample of 90 test taker will provide a sample mean test score within 10 points of the population mean of 515 on the mathematics part of the test? It ranges in ±10 515 + 10 = 525 515...