Answer:
the probability that the sample mean exceeds the W.H.O. minimum = 0.0059
Step-by-step explanation:
Given -
The World Health Organization's (W.H.O.) recommended daily minimum of calories is 2600 per individual .
Mean [tex](\nu )[/tex] = 2460
standard deviation [tex](\sigma )[/tex] = 500
Sample size ( n ) = 81
Let [tex]\overline{X}[/tex] be the sample mean
standard deviation of sample mean = [tex]\sigma _\overline{X} = \frac{\sigma }{\sqrt{n}}[/tex] = [tex]\frac{500 }{\sqrt{81}}[/tex] = 55.55
the probability that the sample mean exceeds the W.H.O. minimum =
[tex]P(\overline{X} > 2600)[/tex] = [tex]P(\frac{\overline{X} - \nu }{\sigma _{\overline{X}}} > \frac{2600 - 2460}{55.55})[/tex]
= [tex]P(Z > 2.52)[/tex]
= [tex]1 - P(Z \leq 2.52)[/tex]
= 1 -.9941
= 0.0059
