Question

Suppose the length of textbooks in a library follows a bimodal distribution with a little right skewness (very mild). The mean of this distribution is 512 pages with a standard deviation of 390 pages.

For each of the following i) draw a picture. ii) label the picture with 2 axes (underneath). iii) label the shorthand for the new distribution. iv) Find the z-score. v) Find the answer.  

a1) What is the probability that a random sample of 36 textbooks has an average of 445.2 pages or less?

Answer 1

Answer 2

Calculation:

1. Understand the Problem:
   We are given a bimodal distribution of textbook lengths with:

• Mean ($\mu$) = 512 pages

• Standard deviation ($\sigma$) = 390 pages
  We are sampling $n=36$ textbooks and want to find the probability that the sample mean ($\bar{X}$) is ≤ 445.2 pages.

• 
  

2. Central Limit Theorem (CLT):
   Since the sample size is large ($n=36$), the sampling distribution of the sample mean will be approximately normal, even though the original distribution is bimodal.

• Mean of sampling distribution (${\mu }_{\bar{X}}$) = $\mu =512$

• Standard error (${\sigma }_{\bar{X}}$) = $\frac{\sigma }{\sqrt{n}}=\frac{390}{\sqrt{36}}=65$

3. Calculate the Z-Score:
   The z-score converts the sample mean to a standard normal value:

   $$z=\frac{\bar{X}-{\mu }_{\bar{X}}}{{\sigma }_{\bar{X}}}=\frac{445.2-512}{65}=\frac{-66.8}{65}\approx -1.0277$$

5. Find the Probability:
   Using the standard normal table (or calculator), the probability corresponding to $z=-1.0277$ is approximately 0.1539 or 15.39%.

6. 
   

7. Conclusion:
   There is a 15.39% chance that a random sample of 36 textbooks will have an average length of 445.2 pages or less .