Null hypothesis "H_0:\\mu=5."
Alternative hypothesis "H_a:\\mu\\ne5."
Test statistic: "z=\\frac{\\bar x-\\mu}{\\sqrt{\\frac{\\sigma^2}{n}}}=\\frac{4.45-5}{\\sqrt{\\frac{4}{900}}}=-8.25."
P-value: "p=2P(Z<-8.25)<0.0001."
Since the P-value is less than 0.05, reject the null hypothesis.
It cannot be reasonably regarded as a sample from a large population whose
mean is 5 cms and varaince is 4 cms.
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