Discussion
The mean height of 10000 children of age 6 years is 41.26" and the standard deviation is 2.24". Then the odds against the possibility that the mean of a random sample of 100 is greater than 41.7" is :
Answer: B
The odds against are 39 : 1.
The sample mean of 100 children has standard error \(2.24/\sqrt{100} = 0.224\). Standardise the value 41.7:
\[ \begin{aligned} z &= \frac{41.7 - 41.26}{0.224} \\ &= \frac{0.44}{0.224} \approx 1.96 \end{aligned} \]
So \(P(\bar{X} \gt 41.7) = P(Z \gt 1.96) \approx 0.025\), and the probability of the opposite event is 0.975. The odds against the event are
\[ 0.975 : 0.025 = 39 : 1 \]
1 : 39 gives the odds in favour, not against. 40 : 1 and 1 : 40 compare the event with the whole, 1 in 40, rather than with its complement.
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