Given `mu=90,sigma=35` :

(a) Find `P(x>100)`

Convert the datum to a standard score: `z=(100-90)/35=.29`

Then `P(x>100)=P(z>.29)` . From the standard normal table we find the area to the right of z=.29 to be .3859

**So `P(x>100)~~.3859` . It happens about 39% ofthe time, so not too unusual.**

(b) For a...

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Given `mu=90,sigma=35` :

(a) Find `P(x>100)`

Convert the datum to a standard score: `z=(100-90)/35=.29`

Then `P(x>100)=P(z>.29)` . From the standard normal table we find the area to the right of z=.29 to be .3859

**So `P(x>100)~~.3859` . It happens about 39% ofthe time, so not too unusual.**

(b) For a sample of `n=10` the sampling distribution has `bar(x)=90,s=35/sqrt(10)~~11.07`

(c) For a sample of size 10 find `P(bar(x)>100)`

Again we convert to a standardized score: `z=(100-90)/(35/sqrt(10))~~.9035`

(Note the use of the standard error)

`P(bar(x)>100)=P(z>.9035)` .The area under the standard normal curve to the right of z=.90 is .1841 (from a standard normal table)

Thus `P(bar(x)>100)~~.1841`

(d) For n=25 find `P(bar(x)>100)`

`z=(100-90)/(35/sqrt(25))~~1.43`

`P(bar(x)>100)=P(z>1.43)=.0764`

(e) Since the sample in (d) is larger, the standard error is smaller. As the sample size increases, the means of the samples get closer to the true population mean (the error decreases).