Question

2) A sample of ?=24 observations is drawn from a normal population with ?=1000 and ?=240. Find each of the following:

A. ?(?¯>1097)

Probability =

B. ?(?¯<906)

Probability =

C. ?(?¯>990)

Probability =

Answer 1

sample size (n) = 24 < 30...but the sample is from a normal population..so, the sampling distribution of the sample mean will follow normal distribution with :-

$mean (\mu_{\bar{x}})=\mu=1000$

$sd (\sigma_{\bar{x}})=\frac{\sigma}{\sqrt{n}}=\frac{240}{\sqrt{24}}=48.9898$

a).$\mathbf{P(\bar{X}>1097)}$

$=P(\frac{\bar{X}-1000}{48.9898}>\frac{1097-1000}{48.9898})$

$=P(Z>1.98)$

$=1-P(Z<1.98)$

$=1-\Phi(1.98)$

$=1-0.9761$[ using standard normal table ]

$=\mathbf{0.0239}$

b).$\mathbf{P(\bar{X}<906)}$

$=P(\frac{\bar{X}-1000}{48.9898}<\frac{906-1000}{48.9898})$

$=P(Z<-1.9188)$

$=\Phi(-1.9188)$

$=\mathbf{0.0275}$ [ in any blank cell of excel type =NORMSDIST(-1.9188) ]

c).$\mathbf{P(\bar{X}>990)}$

$=P(\frac{\bar{X}-1000}{48.9898}>\frac{990-1000}{48.9898})$

$=P(Z>-0.2041)$

$=1-P(Z<-0.2041)$

$=1-\Phi(-0.2041)$

$=1-0.4191$

$=\mathbf{0.5809}$ [ in any blank cell of excel type =NORMSDIST(-0.2041) ]

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