Question

The average grade in a statistics course has been 76 with a standard deviation of 11. If a random sample of 56 is selected from this population, what is the probability that the average grade is more than 80? Use Appendix B.1 for the z-values. (Round your z-value to 2 decimal places and the final answer to 4 decimal places.)

Probability           

Answer 1

Solution :

Given that ,

mean = $\mu$ = 76

standard deviation = $\sigma$ = 11

$\sigma$$\bar x$ = $\sigma$ / $\sqrt$ n = 11 / $\sqrt$ 56 = 1.4699

P($\bar x$ > 80) = 1 - P($\bar x$ < )80

= 1 - P[($\bar x$ - $\mu$ $\bar x$ ) / $\sigma$ $\bar x$ < (80 - 76) / 1.4699]

= 1 - P(z < 2.72)

= 1 - 0.9967

= 0.0033

Probability = 0.0033