Question

please help! stabdard error!

The weight of people in Ozark, Arkansas is normally distributed with a population mean of 185 pounds with a population standard deviation of 25 pounds. 1. What is the standard error of the mean for a random sample of 16 people from Ozark, Ark? 2. How many people must be sampled to decrease the standard error of the mean to 5? 3. What is the probability of the average weight of 16 people in Ozark picked at random will be less than 200 lbs.? 4 There is a raft that takes people across a river that cuts through the middle of town. A warning sign for the raft states "Maximum capacity 3,200 pounds or 16 persons". What is the probability that a random sample of 16 persons from Ozark will exceed the weight limit of 3.200 pounds? (Hint: The raft can carry at most 3,200 lbs. or 16 people. Use this fact.)

Answer 1

3. We ned the probability that average weight I of 16 people will be less than 200 lbs. ie PCX2200) .. y is the sampling distribution of the mean of random samples of & of size n from the central limit theorem i follows a normal distribution with mean ll & standard deviation pla n . 16 al 25 28 2 6.25 Now PCX 600) Find z-score of x=200 & Z. Pall - 200-185 - 15 & 2.40 ? = 6-25 6.25 Now, from the standard normal table, find the p-value for z-score of 2.40 ie PLZ 2208) Rezero) - 0.9918 (from 2-table] . Plan 200) = P(ZL240) - 0.9918 .. & maximum capacity of raft - 3200lbs cors 16 people in If average weight of 16 people – 3200 - 200 We need the probability that 16 people (randomly gelected) will exceed the weight limit of 3200lbs We need PC200) {Wie > 3200) Ta is a 200 but PCT >200) si pere 2007 o 1-0.99182 00082 . Required probability 2 0.0082

The central limit theorem states that the sample mean, $\bar{X}$ follows approximately the normal distribution with mean μ and standard deviation $\sigma$/√n , where µ and $\sigma$ are the mean and standard deviation of the population from where the sample was selected. The sample size has to be large(n≥30) if the population, X is non-normal but if the population, X is normal then regardless of the sample size the sample mean, $\bar{X}$ follows a normal distribution.

In short, $\bar{X}$~N(µ,$\sigma$/√n) if X is normal and if X is non-normal then for n≥30