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2. A random sample of 100 observations from a normally distributed population possesses a mean equal to 83.2 and a standard deviation equal to 6.4. (1) Find a 95% confidence interval for .

Respuesta :

Answer: 95% confidence interval would be (81.95,84.45).

Step-by-step explanation:

Since we have given that

n = 100

mean = 83.2

Standard deviation = 6.4

So, At 95% confidence interval, z = 1.96

so, interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=83.2\pm 1.96\times \dfrac{6.4}{\sqrt{100}}\\\\=83.2\pm 1.2544\\\\=(83.2-1.2544,83.2+1.2544)\\\\=(81.95,84.45)[/tex]

Hence, 95% confidence interval would be (81.95,84.45).

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