# Binomial Distribution Probability question

I am trying to work through a homework problem and am struggling, was hoping you could provide assistance helping me work through the problem.

I am given a world problem with this being the important info:

p = .55
n = 10
find P(X <= 2)


I am trying to figure out how to solve this and this is what I did.

I found the mean using n*p so the mean is 5.5.

I found the standard deviation using Sqrt(np(1-p)) and found that it is 1.573.

I then assumed the next step would be to calculate a z-score for this: I did

(2-5.5) / 1.573 = -2.23


I then assumed that the answer to this problem would be the P(Z <= -2.23) which, by using a z-score table I found to be 0.0129; however apparently the correct answer is 0.0045.

Where am I going wrong in my process? Thank you very much for your assistance.

• $n=10$ is not sufficiently high to use normal approximation with high accuracy. Find probabilities $P(X=0)$ and $P(X=1)$ from probability mass function and add. – NCh Apr 14 '17 at 2:14
• @NCh What exactly is the probability mass function? I apologize I am quite an amateur when it comes to stat. – Biggytiny Apr 14 '17 at 2:15
• en.wikipedia.org/wiki/… – NCh Apr 14 '17 at 2:16
• @NCh thank you very much, I appreciate it! – Biggytiny Apr 14 '17 at 2:17
• @NCh my mistake, I got it now. Calculation error, thank you! Can you explain why this is the correct method to approach this? I do not quite undrestand. – Biggytiny Apr 14 '17 at 2:22

With $n = 10$ and $p = 0.55,$ you're just at the edge of what many textbooks consider suitable for a normal approximation. It helps that $p$ is not far from 1/2.

For $X \sim \mathsf{Binom}(10,.55),$ the exact value is $$P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)\\ = {10 \choose 0}(.45)^{10} + {10 \choose 1}(.55)(.45)^9 + {10 \choose 2}(.45)^8(.55)^2,$$ which is not too difficult to compute on a hand calculator. In R statistical software, this exact value is found to be 0.0274, correct to four places. This is in poor agreement with the 'correct' answer you were given.

pbinom(2, 10, .55)
## 0.02739184


Perhaps the answer key had in mind that you would use a normal approximation. Then $\mu = E(X) = np = 5.5$ and $\sigma = SD(X) = \sqrt{np(1-p)} = 1.573213,$ as you say. Then

$$P(X \le 2) = P(X \le 2.5) = P\left(\frac{X-\mu}{\sigma} \le \frac{2.5-5.5}{1.5732} \right) \approx P(Z \le - 1.91) = 0.0281,$$

which agrees with the exact value to two places (all that can be expected using a normal approximation, especially one on the borderline), but this still does not agree with answerbook's value 0.0045. Notice that I used 2.5 instead of 2; this is called the 'continuity correction' which is probably explained in your text.

Using your value $z = \frac{2-5.5}{1.573} = -2.225,$ without the continuity correction, I get $P(Z \le -2.225) = 0.0130$ (with interpolation in the table), in substantial agreement with your answer.

In conclusion, using the normal approximation without continuity correction, your answer seems OK, and I can find no support for the 'supposedly correct' answer. I think you can see that it is worthwhile learning about the 'continuity correction', if it is covered in your book. In this era of readily available computation, I would not use a normal approximation (especially one near the borderline) for a serious application of the binomial distribution.

In the sketch below, the heights of the blue bars to the left of the vertical red dotted line represent the exact value of $P(X \le 2).$ [The 'bar' for $P(X=0)$ is so small it doesn't show at the scale of the sketch.]

According to the continuity correction, $P(X \le 2)$ corresponds roughly with the area under the approximating normal curve to the left of the dotted line. You can see that the normal curve is not a bad approximation in the left tail (near 0) of the binomial PDF. [The area under the normal density curve between 1.5 and 2.5 corresponds approximately to $P(X = 2).$ By omitting the area between 2.0 and 2.5, without continuity correction, you are missing more than half of $P(X = 2).$]