Thread: The mean height of 10,000

The mean height of 10,000 children of age 6 years is 41.26 and standard deviation is 2.24.Find the odds against the possibility that the mean of random sample of 100 is greater than 41.7
Solution:

n_1 = 10,000
\bar x = 41.26
s_1 = 2.24

n_2 = 100

I don't know what should be the hypothesis .........

Originally Posted by Athena

The mean height of 10,000 children of age 6 years is 41.26 and standard deviation is 2.24.Find the odds against the possibility that the mean of random sample of 100 is greater than 41.7
Solution:

n_1 = 10,000
\bar x = 41.26
s_1 = 2.24

n_2 = 100

I don't know what should be the hypothesis .........

The population is the 10000 children, so the mean and standard deviation is 41.26 and 2.24 (there is no approximation involved).

The distribution of the sample mean of a sample of 100 is approximately normal with mean 41.26 and standard deviation 0.224 (the population SD divided by the square root of the sample size). This is approximate as it is treating the sample as though it is sampled with replacement.

So the z-score of a mean of 41.7 is (41.7-41.26)/0.224, and the required odds of seeing a z-score less than this are:

P(z<(41.7-41.26)/0.224)/(1-P(z<(41.7-41.26)/0.224))

where P(.) is the cumulative standard normal probability function.

CB