Question

Question 12 Current research indicates that the distribution of the life expectancies of a certain protozoan is normal with a mean of 49 days and a standard deviation of 10.1 days. Find the probability that a simple random sample of 25 protozoa will have a mean life expectancy of 54 or more days. 0.2067 b) O0.0067 0.6897 0.9933 e) 0.0133

Answer 1

12)

Solution :

Given that ,

mean = $\mu$ = 49

standard deviation = $\sigma$ = 10.1

n = 25

$\mu$_{$\bar x$} = 49 and

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 10.1 / $\sqrt$ 25 = 2.02

P($\bar x$? $\geq$ 54) = 1 - P($\bar x$$\leq$ 54) = P(($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$}$\leq$ (54 - 49) / 2.02) = 1 - P(z $\leq$ 2.473)

    Using standard normal table,

P($\bar x$$\geq$ 54) = 1 - 0.9933 = 0.0067

Probability = 0.0067

Option b) is correct .