Question

Perform a continuity correction to rewrite the probability P(100 ? x < 115) that involves a discrete random variable (x) as a probability involving a continuous random variable (x').a. P(100.5 < x' < 115.5)

b. P(99.5 < x' < 115.5)c. P(100.5 < x' < 114.5)
d. P(99.5 < x' < 114.5)

Answer 1

Option B

Explanation:

Given:

- P ( 100 <= x' <= 115 )

Find:

- The correct expression for continuity correction:

Solution:

- For continuity correction we will observe the signs for each limit '' < " , " > " , " < = " , or " = > ".

- The given expressions has the sign " < = " and " = > ". For continuity correction i.e required for binomial distribution approximation with normal distribution. Since normal distribution can not compute probability " = " or any equals to sign. We convert the " = " sign to equivalent " < " or " >" signs.

- So, to convert " 100 < = x' " into " a < x' ". We can write that x > 99.

Then we find and average between the two i.e 99.5. So correction for LHS of expression is (99.5 < = x').

- Similarly for RHS we have " x' <= 115 " we need to convert it into " x' < a ".

We can write it as x'< 116. Now again take average of the two numbers Left and right we have 115.5. Hence, (x' < 155.5)

- The final expression is: P(99.5 < x' < 155.5) @ which normal approximation is to be performed. Option B