This question comes from Probability and Statistics for Engineering and the Sciences 9th Ed. by Devore, Jay Can you explain what a continuity correction is and how it is calculated? Also can you explain why .5 was chosen for this particular problem? Will it always be subtracted from one end and added to the other? Thank you for your time.
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
This question comes from
Can you explain what a continuity correction is and how it is calculated?
Also can you explain why .5 was chosen for this particular problem?
Will it always be subtracted from one end and added to the other?
Thank you for your time.
![The random variable \( X \) is originally discrete in nature. The approximate distribution of \( X \) is normal, which is continuous in nature. As a discrete variable is being approximated by a continuous variable, a continuity correction is made by subtracting 0.5 from the lower limit 35 and adding 0.5 to the upper limit 70.
Thus,
\[
P(35 \leq X \leq 70) \approx P(34.5 \leq X \leq 70.5)
\]
\[
= P\left(\frac{34.5 - \mu}{\sigma} \leq \frac{X - \mu}{\sigma} \leq \frac{70.5 - \mu}{\sigma}\right) \quad \text{(by standardizing)}
\]
\[
= P\left(\frac{34.5 - 50}{\sqrt{50}} \leq Z \leq \frac{70.5 - 50}{\sqrt{50}}\right)
\]
\[
= P\left(Z \leq \frac{20.5}{\sqrt{50}}\right) - P\left(Z \leq \frac{-15.5}{\sqrt{50}}\right)
\]
\[
= \Phi(2.90) - \Phi(-2.19)
\]
From Table A.3, “Standard Normal Curve Areas”, we can find the values for \(\Phi\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f9cc468-5045-4ba5-8445-556d55f6f682%2Ffcbde675-f808-45e3-b85b-4f8d4142e914%2Fca76n2e_processed.png&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
