Question

eBook Video A population has a mean of 300 and a standard deviation of 80. Suppose a sample of size 125 is selected and X is used to estimate μ. Use z-table. a. What is the probablity that the sample mean will be within+3 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places) b. What is the probability that the sample mean will be within+/- 19 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places)

Answer 1

1. The population mean is u = 300, and the standard deviation is o = 80. The sample size is n=125. By central limit theorem, the sampling distribution of sample mean is normal with mean u = 300 and the standard deviation 'Th 1125 ö 80-7.1554. a. The probability that the sample mean will be within 3 of the population mean is calculated as follows: II-3-1 Ý- I II+ 3- LAN P(4-3<ă <l+3)=PL (Standardization) on on oln - 3 3 (7.15547 .1554) = P(-0.42<Z<0.42) = P(Z <0.42) - P(Z<-0.42) = 0.6628 -(1-0.6628) (Since P(Z<-0.42)=1-P(Z <0.42)) = 0.3372 (From standard normal table) P(u-3<ă <u+3) = 0.3256 = P - < 7 < b. The probability that the sample mean will be within +19 of the population mean is calculated as follows: P(4-19<ā< ļ+19) = P1“ U-19-\ I-! !+19-! Toln Toltola (Standardi zartion) -1919 7.1554 7.1554 = P(-2.66<Z<2.66) = P(Z < 2.66)- P(Z<-2.66) = 0.9961 – (1-0.9961) (Since P(Z<-2.66) =1 - P(Z < 2.66)) = 0.9922 (From standard normal table) P(4-19<ă <pl+19) = 0.9922

2. In EAI sampling problem, the population mean is u = $51,900 and population standard deviation is o = $5000. The sample size is n=20, there is 0.3453 probability of obtaining a sample mean within +$500 of the population mean. By central limit theorem, the sampling distribution of sample mean is normal with mean o 5000 Il = 51,900 and the standard deviation – Tn Tai a. The probability that the sample mean will be within $500 of the population mean when n= 40 is calculated as follows: plu-500 <i <4+500)=pl -300- X- ,\+500 – 11 Toin olin olun (Standardizarti on) -500 500 ) 5000/40 5000/140) = P(-0.63<Z<0.63) = P(Z <0.63)- P(Z<-0.63) =0.7357 – (1-0.7357) (Since P(Z<-0.63)=1-P(Z <0.63)) = 0.4714 (From standard normal table) P(4-500<ă<+500)= 0.4714 b. The probability that the sample mean will be within $500 of the population mean when n=80 is calculated as follows: P(u-500<I<!+500) = P plu-500- 1 1 - +500 – 4 (Standardization) lolin olin olin (Standardizartion) -500 500 <Z «- 5000/ 180 . 5000/ 80 ) = P(-0.89 <Z<0.89) =P(Z < 0.89) - P(Z<-0.89) = 0.8133-(1–0.8133) (Since P(Z<-0.89)=1- P(Z <0.89)) = 0.6266 (From standard normal table) P(4-500<ā< [1+500)=0.6266