Question

Suppose that you are an elementary school teacher and you are evaluating the reading levels of your students. You find an individual that reads 60.3 word per minute. You do some research and determine that the reading rates for their grade level are normally distributed with a mean of 100 words per minute and a standard deviation of 23 words per minute.

a. At what percentile is the child's reading level (round final answer to one decimal place).

b. Create a graph with a normal curve that illustrates the problem.

For the graph do NOT make an empirical rule graph, just include the mean and the mark off the area that corresponds to the student's percentile.

c. Make an argument to the parents of the child for the need for remediation. Structure your essay as follows:

1. A basic explanation of the normal distribution
2. Why the normal distribution might apply to this situation
3. Describe the specific normal distribution for this situation (give the mean and standard deviation)
4. Interpret the answer to part a. including a definition of percentile.
5. Explain how the graph created in part b. represents the child's reading level.
6. Use the answers to parts a. and b. to emphasize the gravity of the situation.
7. Give a suggested course of action.

Answer 1

Let the random variable representing reading rates be X. Given, X ~ N(100, 23²)

a) Let the child's reading level be at percentile p

Hence, we have

p = P(X <= 60.3)

= P(Z <= (60.3 - 100) / 23 ) = P( Z <= -1.726) = 0.0418

Hence, child's reading level is at 4.18 percentile.

b) The normal curve below shows the position where the child lies among the students at the school

c)  

1) Normal distribution is a bell-shaped symmetrical probability distribution as shown above.

It can be assumed to be applicable in this situation as the sample size is quite large. This can also be verified by drawing a histogram of the reading rates of the students.

The normal distribution for this situation as explained above is N(100, 23²), i.e. has a mean of 100 and std deviation of 23

The answer to part a, 4.18 percentile, means the student is at 4.18 percentile in the school in terms of reading rates, i.e. only 100 - 4.18 = 95.82 percentile of the students have a reading rate better than this student

The graph created in part b shows the point on the X-axis (set of possible values of reading rate) where the reading rate of the student lies and the probability of a student having a reading rate lesser than his (0.0418 in this case)

The student's reading rate lies at 4.18 percentile, i.e. there are 100 - 4.18 = 95.82 percent students having better reading rate than him, which is nearly the entire population of students. Hence, serious improvement is required to improve his reading rate.

There is a serious need to intervene to improve the reading rate of the student by conducting dedicated reading sessions for him/her. The near-zero percentile of reading rate suggests this might be one of the most important areas to focus for the student.