Question

A library wants to determine the effectiveness of their summer literacy program among low-income children. Because surveying the large numbers of students in the program would require too many resources the library staff interviews 30 randomly chosen children among the low-income program attendees. The 30 sampled children are given a reading test before and after the program.A) The difference in the reading test scores (after – before) has mean 10 and standard deviation 4. Assuming the score differences are normally distributed, what percent of the children showed any improvement (difference > 0) in reading ability?B) What percent of children improved by more than 15 points?

Answer 1

Answer:

(A) P (D > 0) = 99.38%

(B) P (D > 15) = 10.56%

Step-by-step explanation:

The random variable D = difference, is defined as the difference between the reading test scores after and before the program.

The random variable D follows a normal distribution with mean, $\mu_{D}=10$ and standard deviation, $\sigma_{D}=4$.

(A)

Compute the probability that the children showed any improvement, i.e.

P (D > 0):

$P(D>0)=P(\frac{D-\mu_{D}}{\sigma_{D}} >\frac{0-10}{4} )=P(Z>-2.5)=P(Z<2.5)$

Use the standard normal random variable to determine the probability.

$P(D>0)=P(Z<2.5)=0.9938$

The percentage of children showed any improvement is:

0.9938 × 100 = 99.38%

Thus, 99.38% of children showed improvement.

(B)

Compute the probability that the children improved by more than 15 points, i.e. P (D > 15):

$P(D>15)=P(\frac{D-\mu_{D}}{\sigma_{D}} >\frac{15-10}{4} )=P(Z>1.25)=1-P(Z<1.25)$

Use the standard normal random variable to determine the probability.

$P(D>0)=1-P(Z<1.25)=1-0.8944=0.1056$

The percentage of children improved by more than 15 points is:

0.1056 × 100 = 10.56%

Thus, 10.56% of children showed improvement by more than 15 points.