Question

f. Answer the same question 5.2. The BOD (biological oxygen demand) in a stream has the mean 80 mg/l and standard deviation 16 mg/1. What is the probability that the annual mean BOD to be computed from 12 monthly samples will exceed 90 mg/? a. Use the asymptotic sampling distribution. b. Use the exact sampling distribution.

I want part b because I have some problems with t distribution

Answer 1

In part a of this problem we use the asymptotic distribution of the annual mean BOD which is Standard Normal distribution and in the part b we use the exact distribution of the sample mean BOD , since the population standard deviation is estimated by sample sd. To calculate the probability we take help of standard normal table and the function T.dist in Excel to get the left tailed probabilities and subtracting them from 1 in part a and b respectively.The step by step explanations are given in the images....

bet x,,ta... be 12 monthly samples. drawn from the stream. We assume BOD - Normal diste Then x,, tag, X12 WN (8082) - variance of Bol of the stream which is unknown. Parta The assemolate chotelution of annual mean BOD se & is 7-80_213. (X-80) NO) where sis the same forlard deviation. Henathe probalihity that the oromal neon BOD is greater than googll is PLT 790] - PLX-8090-807 V821312 Nome in this problem 8 = 16 - PL * >903 – PS2V5 (20), 213410] - P[21368-90), 2.165]. -|- (2-165) - 0.0153 [: 28(77891 (0,97 Hence the required probability, 13 0.0153 e have to find the probability $77907 by using erod distrudention. We krom that & N (80, 8). 16 Parble Here we have independently

a win 2-) Hence 288-so) Alists with def 11 1 23.8" 2 213(-80) to which is the exact disbrinution Therefore P[7 790]=P[213 (7-80) > 20165] 21-0.97338 [": 38 (+59) Ang = 0.0266 The probability using the exact distribution, pa 0.0266. herefore 13 16