Question

The Mental Development Index (MDI) for infant development is a standardized measure used in longitudinal

follow-up of high risk infants. The MDI scores are known to be normally distributed with mean 100 and variance
256. For a study of infants, a random sample of size 400 is selected.

Answer 1

Answer on Question #63259 – Math – Statistics and Probability

Question

The Mental Development Index (MDI) for infant development is a standardized measure used in longitudinal follow-up of high risk infants. The MDI scores are known to be normally distributed with mean 100 and variance 256. For a study of infants, a random sample of size 400 is selected.

(a) find the standard error of the mean MDI originating from this sample.

(b) find the percentage of the mean MDI score from this sample that

(i) will be greater than 101.6;

(ii) will be less than 98.4;

(iii) will be between 98.4 and 101.6.

Solution

(a)

$$SE = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{256}}{\sqrt{400}} = 0.8.$$

(b)

(i)

$$P(\bar{x} > 101.6) = P\left(Z > \frac{101.6 - 100}{0.8}\right) = P(Z > 2) = 1 - P(Z < 2).$$

From z-table

$$P(Z < 2) = 0.9772;$$ $$P(\bar{x} > 101.6) = 1 - 0.9772 = 0.0228.$$

The percentage is 2.28%.

(ii)

$$P(\bar{x} < 98.4) = P\left(Z < \frac{98.4 - 100}{0.8}\right) = P(Z < -2) = P(Z > 2) = 0.0228.$$

The percentage is 2.28%.

(iii)

$$\begin{array}{l} P(98.4 < \bar{x} < 101.6) = P\left(\frac{98.4 - 100}{0.8} < Z < \frac{101.6 - 100}{0.8}\right) = P(-2 < Z < 2) = \\ = P(Z < 2) - P(Z < -2) = = 1 - P(Z > 2) - P(Z < -2) = \\ = 1 - 0.0228 - 0.0228 = 0.9544. \end{array}$$

The percentage is 95.44%.

Answer: (a) 0.8; (b) (i) 2.28%; (ii) 2.28%; (iii) 95.44%.

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