The Mental Development Index (MDI) of the Bayley Scales of Infant Development is a standardized measure used in longitudinal follow-up of high-risk infants. The scores on the MDI have approximately a normal distribution with a mean of 100 and standard deviation of 16. We are going to randomly select 64 children and average their MDI scores. What is the probability that the average is under 104?

Solution

Let $X$ denote the Mental Development Index score.

The scores on the MDI have approximately a normal distribution with a mean of 100 and standard deviation of 16.

That is $X\sim N(\mu, \sigma^2)$, where $\mu = 100$, $\sigma = 16$.

A sample of $n = 64$ children are selected at random, then $\overline{X} \sim N(\mu, \sigma^2/n)$.

So $Z=\dfrac{\overline{X}-\mu}{\sigma/\sqrt{n}} \sim N(0,1)$ standard normal variate.

The probability that the average MDI score is under 104 is

$$ \begin{aligned} P(\overline{X} <104) & = P\bigg(\frac{\overline{X} -\mu}{\sigma/\sqrt{n}} < \frac{104 -100}{16/\sqrt{64}}\bigg)\\ &= P(Z < 2)\\ &= 0.9772 \end{aligned} $$

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