Tute 4 - finding the CLT and sample size PDF

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finding the CLT and sample size...

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MAS183 Tutorial 4

1a) The central limit theorem states that if you have a population with mean μ= 650,000 and standard deviation σ= 450,000, The central limit theorem states is a random samples from the population of 45 randomly- selected Perth Houses, then the distribution of the sample means it would be a normally distributed. The mean μ x = μ= $650,000 The standard deviation σ= σ/√n = 67082.04 1b) The distribution of day – number in samples will also follow uniform distributed, because sample size is less than 30. We cannot apply central limit theorem

1c) Here sample size is small, but it is given that population is normally distributed. According to central limit theorem, sampling distribution of the average systolic blood pressure in a sample of 15 people from this population will be approximately normal distribution 2a) X= Serum cholesterol levels   x = Sample mean cholesterol levels. Given that X=N (170,30) Therefore, x =N (170, 30/√n ), N= Sample size (a) x = 180 mg/dL in a sample of size 15. (a) x = 180 mg/dL in a sample of size 15.   x =N (170, 30/√15),

P(X(180-170)/ (30/√15)) P(X1.291) P(X 180)= 0(2>(180-170)/ (30/√15)) P( x> 180)= (2>1.291) P( x> 180)= 0.09853 Therefore, using standard; the observation is not usually large in random samples of size 15 from this population. (b) x = 180 mg/dL in a sample of size 60. (b)  x = 180 mg/dL in a sample of size 60. x =N (170, 30/√60), P(X(180-170)/ (30/√60)) P(X2.582) P(X 180)= 0(2>(180-170)/ (30/√60)) P( x> 180)= (2>2.582) P(x> 1 P( x> 180)= 0(2>(180-170)/ (30/√60)) P( x> 180)= (2>2.582) P(x> 1 P( x> 180)= 0(2>(180-170)/ (30/√60)) P( x> 180)= (2>2.582) P( x> 180)= 0.00494 This is a random sample size of 60. This population as observed, would show usually large valves. Q3a) Given, μ= 58, σ=27 and n=500. They are positively skewed distribution. Q3b) The distribution of internet access fees is skewed, with most householdhaving cheaper plans and few playing much more The distribution of internet access fees is skewed, with most household having cheaper plans and few playing much more. Q3c) P (the mean monthly access fee in the sample would be less than $55). Q3d) ans and few playing much mor The distribution of internet access fees is skewed, with most householdhaving cheaper plans and few playing much more The distribution of internet access fees is skewed, with most householdhaving cheaper plans and few playing much more The distribution of internet access fees is skewed, with most householdhaving cheaper plans and few playing much mo 4a)

(a) Normal, with μ = .1 and σ = 01732. 300 9.1.)1( = × = − n pp (a) Normal, with μ = .1 and σ = 01732. 300 9.1.)1( = × = − n pp (a) Normal, with μ = .1 and σ = 01732. 300 9.1.)1( = × = − n pp Normal, with μ = .1 and σ = 01732

5)See unit guide pg. 96- 97. ñ = 643 + 4 = 647

pD = 74+2 647 = .1175 pD = 74+2 647 = .1175 We can be 95% sure that between 9.3% and 14.2% of people in this population are left-handed We can be 95% sure that between 9.3% and 14.2% of people in this population are left-handed

See unit guide page 94- 95...