Approximation with a normal distribution

Every day Alice tries the stroke playing tennis until she reaches $50$ strokes. If each stroke is good with a probability of $0,4$, independently from others, approximately what is the probability that at least $100$ attemps are necessary to success?

Let X be a binomial random variable with parameters $$n=100$$ and $$p=0,4$$

Since n is large we can approximate with a normal distribution with parameters $$\mu=40$$ and $$\sigma=\sqrt{0,4*100*0,6}=\sqrt{24}$$

Appling the normal approximation \begin{align}P(X> 49,5)&=1-P(x<49,5)\\&=1-P((X-\mu )/\sigma < (49,5-40 )/ \sqrt{24} )\\&=1-P((X-\mu )/\sigma < 1,939 )\\&=1-\Phi (1,939)\\&=1-0,9737\\&=0,0263\end{align}

But the solution on the book is $0,974$ (it could be $P(x<49,5)$).

You have a couple of answers using different methods, so this is to show that they are both correct approaches (+1 and +1). I do not see that any method leads to exactly the answer 0.974, but the discrepancy may have to do with the method of approximation used.

By thinking the problem through carefully, you can work the problem using either the binomial or the negative binomial distribution.

Binomial. Following @SiongThyeGoh's logic, if $X$ is the number of successes in $n = 99$ tries, and you have $X \le 49,$ then you will need at least 100 tries to get 50 successes. So we have $X \sim Binom(99, .4)$ and we seek $P(X \le 49).$

Using R statistical software, the statement pbinom(49, 99, .4) returns $0.9781,$ rounded to four places.

If you use the normal approximation with $\mu = np = 99(.4) = 39.6$ and $\sigma = 4.8944,$ then you have $$P(X \le 49) = P(X \le 49.5) \approx P\left(Z < \frac{49.5-39.6}{4.8744} = 2.03\right),$$ where $Z$ is standard normal. Printed tables give $P(Z \le 2.03) = .9788.$ You can avoid some round-off error in R, using pnorm(49.5, 39.6, 4.8744) to get $0.9789.$

Negative Binomial (counting failures) There are several parameterizations of the negative binomial distribution. One of them defines $Y = 0, 1, 2, \dots$ to be the number of failures occurring before the $r$th success. You would need at least 50 failures before the 50th success in order to require 100 or more tries.

This is the version of the negative binomial distribution implemented in R. So we need $P(Y < 50) = P(Y \le 49),$ which can be found in R with the statement pnbinom(49, 50, .6) to get $0.9781,$ rounded to four places.

 x = 0:70;  pdf = dnbinom(x, 50, .6)
plot(x, pdf, type="h", lwd=3, col="blue", main="PDF of Negative Binomial Dist'n Counting Failures")
abline(v=49.5, col="red", lwd=2, lty="dotted")
abline(h=0, col="green2")


Negative Binomial (counting trials). In @heropup's Answer, the negative binomial random variable $X$ is the number of trials, not failures, required to get $X$ successes, where the probability of success on any one trial is 0.4. You can use a normal approximation with her $\mu$ and $\sigma$ in the R statement 1 - pnorm(99.5, 125, 13.693) to get $0.9687,$ rounded to four places.

Ordinarily, you cannot depend on more than about two decimal places of accuracy form a normal approximation to the binomial or negative binomial distribution. You can see what you get using normal tables.

The illustration below shows the skewness of this negative binomial distribution and the less-than-ideal normal approximation. Here, you are interested in the probability to the right of the vertical red line.

mu = r/p;  sg = sqrt(r/p^2);  sg.r = round(sg,2) # for normal aprx
head=paste("PDF of Neg Binomial Dist'n Counting Trials with NORM(",mu,",",sg.r,") Density", sep="")
x = 50:170;  r = 50; p=.4; q= 1-.4  # neg bin parameters
pdf = choose(x-1,r-1)*p^r*q^(x-r)   # this pdf not programmed in R
plot(x, pdf, type="h", lwd=3, col="blue", main=head)
abline(v=99.5, col="red", lwd=2, lty="dotted")
curve(dnorm(x, 125, 13.7), add=T);  abline(h=0, col="green2")


If you are going to use the negative binomial distribution, you should look to see what version your text or lectures are using, and learn that.

If $Y \sim Bin(99, 0.4)$

$Y < 50$ implies that there are less than $50$ succeses in $99$ trials, and hence at least $100$ attempts are necessary to reach $50$ succeses.

Hence the quantity that you are suppose to compute is $P(Y<50)$

You are using the wrong model for the event of interest. Specifically, you are interested in the number of attempts needed to achieve a certain pre-defined number of successes, not the number of successes observed in a pre-defined number of attempts.

Therefore, the model you should use is not binomial, but negative binomial. The random number of attempts $X$ needed to observe $r = 50$ successes, when each success has independent probability of $p = 0.4$ of occurring, is a negative binomial random variable with mean $$\mu = \frac{r}{p} = 125,$$ and variance $$\sigma^2 = \frac{(1-p)r}{p^2} = \frac{375}{2}.$$ Therefore, your normal approximation of $X$ should be based on these moments, not the ones you specified.

• "the model you should use is not binomial, but negative binomial" Binomial works like a charm here: the number of attempts to get K=50 successes is larger than N=100 if and only if the number of successes during the N=100 first attempts is smaller than K=50. The number of attempts to get K successes follows a negative binomial distribution and the number of successes during the N first attempts follows a binomial distribution. – Did Dec 30 '16 at 12:09

Considering an approximation with a normal distribution of the binomial random variable, I think that we should calculate $$P(X<49.5)$$ because the number of attempts must be more than 100.

when n is large, a binomial random variable with parameters n and p will have approximately the same distribution as a normal random variable with the same mean and variance as the binomial. If the random variable is negative binomial is possible to apply this approximantion?

• I am not sure just how well and for what choices of $n$ and $p$ the negative binomial distribution is well approximated by the normal (see the last part of my Answer). Certainly, the normal approximation to the binomial works well. – BruceET Dec 30 '16 at 9:17