A population has a mean of 300 and a standard deviation of 90. Suppose a sample of size 100 is selected and x is used to estimate μ.

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

b. What is the probability that the sample mean will be within +/- 14 of the population mean (to 4 decimals)?

Respuesta :

We use the formula: [tex]z = \frac{x-\mu}{\frac{\sigma}{ \sqrt{n}}} = \frac{x-\mu}{\frac{90}{\sqrt{100}}} = \frac{x-\mu}{9}[/tex]
a. [tex]z = \frac{5}{9} = 0.56[/tex], so we then find P(-0.56 < z < 0.56) from z-tables, and this is equivalent to 0.4246.
b.[tex]z = \frac{14}{9} = 1.56[/tex], so we find P(-1.56 < z < 1.56) from z-tables, and this value is equivalent to 0.8812.