Respuesta :
Answer:
The probability is 0.2423.
Step-by-step explanation:
Given mean per capita = 19292 dollars
Given the variance = 540225
Now find the probability that the sample mean will be less than 19269 dollar when the sample is 499.
Below is the calculation:
[tex]\bar{X} \sim N(\mu =19292, \ \sigma = \frac{\sqrt{540225}}{\sqrt{499}}) \\\bar{X} \sim N(\mu =19292, \ \sigma = 32.90) \\\text{therefore the probability is:} \\P (\bar{X}< 19269) \\\text{Convert it to standard normal variable.} \\P(Z< \frac{19269-19292}{32.90}) \\P(Z< - 0.6990) \\\text{Now getting the probability from standard normal table}\\P(Z< -0.6990) = 0.2423[/tex]
