Question

Human body wteeratr"re normaly dsrbuted with a mean of 9820 F and a standard devat on of 082 F. 19people are randonly selected find the p dabity that their mean boytemperatow. be less than 98.50 F. Round your answer to four decimal places OA. 04626 OC 09826 O D. 00833

Answer 1

Solution :

Given that ,

mean = $\mu$ = 98.20

standard deviation = $\sigma$ = 0.62

P(x < 98.50) = P((x - $\mu$) / ($\sigma$/$\sqrt$n) < (98.50 - 98.20) / (0.62/$\sqrt$19) = P(z < 2.11)

Using standard normal table,

P(x < 9.1) = 0.9826

Probability = 0.9826

Option c) is correct .