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.
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.
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 $P(X = 2).$]