Question

The Mental Development Index (MDI) of the Bayley Child Development Scales is a standardized measure used to observe infants over a period of time. It is approximately normal with a mean of 100 and a standard deviation of 16. (Approximately 4 decimal places) to. If 5 children are studied independently, what is the probability that they all have an MDI of at least 100? b. What is the value of the Interquartile Interval (IQR) of the MDI scores?

Answer 1

Answer:

(a) The probability that the five children selected will have an MDI of at least 100 is 0.50.

(b) The IQR of the MDI scores is 21.60.

Step-by-step explanation:

Let the random variable X represent the MDI.

The random variable X follows a Normal distribution with mean, μ = 100 and standard deviation, σ = 16.

(b)

Compute the probability that the five children selected will have an MDI of at least 100 as follows:

$P(\bar X\geq 100)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}\geq \frac{100-100}{16/\sqrt{5}})\\\\=P(Z\geq 0)\\\\=0.50$

Thus, the probability that the five children selected will have an MDI of at least 100 is 0.50.

(b)

The Inter-quartile range of a Normally dstributed data is:

IQR = 1.34896 × σ

       = 1.34896 × 16

       = 21.58336

       ≈ 21.60

Thus, the IQR of the MDI scores is 21.60.