Question

The World Health Organization's (W.H.O.) recommended daily minimum of calories is 2600 per individual. The average number of calories ingested per capita per day for the US is approximately 2460 with a standard deviation of 500. If we take a random sample of 81 individuals from the US, what is the probability that the sample mean exceeds the W.H.O. minimum?

Answer 1

the probability that the sample mean exceeds the W.H.O. minimum = 0.0059

Explanation:

Given -

The World Health Organization's (W.H.O.) recommended daily minimum of calories is 2600 per individual .

Mean
`(\\u )` = 2460

standard deviation
`(\sigma )` = 500

Sample size ( n ) = 81

Let
`\overline{X}` be the sample mean

standard deviation of sample mean =
`\sigma _\overline{X} = (\sigma )/(√(n))` =
`(500 )/(√(81))` = 55.55

the probability that the sample mean exceeds the W.H.O. minimum =

`P(\overline{X} > 2600)` =
`P(\frac{\overline{X} - \\u }{\sigma _{\overline{X}}} > (2600 - 2460)/(55.55))`

=
`P(Z > 2.52)`

=
`1 - P(Z \leq 2.52)`

= 1 -.9941

= 0.0059