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My question is as follows:

A fair die is rolled $120$ times. Find the probability that $5$ is on the top:

a. between $30$ and $40$ times,

b. between $18$ and $50$ times,

c. more than $70$ times.

Hint: Use the approximation of the Binomial distribution to the normal distribution.

I am close to solving it. However, I am stuck, since I obtained a standard deviation value of $4.0825$ and therefore one of the $z$-scores as $4.9$. My solution is as follows:

$x$ = event of having a $5$ shown on top

$P(x) = 1/6$

This is a binomial probability experiment

$np(1-p) \ge 10$?

$120 * \frac{1}{6} ( 1 - \frac{1}{6}) = \frac{50}{3} > 10$

mean $= np = 120$;

standard deviation $= \sqrt{\frac{50}{3}} = 4.0825.$

$X~N(\mu=20, \sigma=4.0825)$

a) $P(30\le x\le 40) =$ ?

$z = \frac{x - \mu} {\sigma} $

$P\left(\frac{30 - 20}{4.0825} \le z \le \frac{40 - 20}{4.0825} \right)= P (2.45\le z\le4.9)$

At this point, I am stuck, since the $p$-value for a $Z$-score of $4.9$ is impossible to find (an online calculator gave me a value of $0.9999995$). Is my solution wrong or should I proceed with the given $p$-value?

Thank you all.

: )

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2 Answers 2

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I wanted to write this in a comment but got carried away and ran out of characters:

Your working all looks pretty fine to me (although you have missed out the end of the line where you work out the mean, but you then do use the correct value ($20$) in your workings). Also I think it's important that typically we write the Normal distribution as $X \sim N(\mu, \sigma^2)$, not $X\sim N(\mu, \sigma)$. Again, you've used the correct values in your work though.

I'm guessing that the low probability is what is making this difficult. Typically to approximate the Binomial to the Normal you want $p \approx \frac 12$. Whereas here obviously we have a much lower $p$.

I think in your shoes I would take the value to be $\approx 1$, and write a small note alongside my answer explaining why I've done this. If this is a homework question then I'm sure your teacher would be more than happy to explain what there thinking is about this.

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  • $\begingroup$ Hi there. Thank you for your answer. When you say "Whereas here obviously we have a much lower p", what do you mean? What p are you referring to? $\endgroup$ Oct 12, 2018 at 10:49
  • $\begingroup$ The same $p$ you refer to: $p$ is the probability that we get a $5$. Which is $\frac{1}{6}$. $\endgroup$
    – MRobinson
    Oct 12, 2018 at 10:54
  • $\begingroup$ oh okay. I see. and I should set p to 1? or what variables value should I set approximately to 1? I am referring to this statement: "I think in your shoes I would take the value to be ≈ 1... " $\endgroup$ Oct 12, 2018 at 11:01
  • $\begingroup$ That's your $p$ value of $4.9$, unfortunately you have $2$ different $p$s in your question! $\endgroup$
    – MRobinson
    Oct 12, 2018 at 11:07
  • $\begingroup$ Thank you for everything. I understand it now. $\endgroup$ Oct 12, 2018 at 11:20
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As a check of your work, the exact probabilities are given as follows:

> sum(dbinom(30:40,120,1/6))
[1] 0.01287915
> sum(dbinom(18:50,120,1/6))
[1] 0.7240251
> sum(dbinom(71:120,120,1/6))
[1] 1.112216e-25

You will notice that these probabilities are slightly higher than the ones you obtained since you are not using a continuity correction. To be more precise, you could take

$$P(30\leq X\leq 40)\approx \Phi\left(\frac{40.5-20}{4.0825}\right)-\Phi\left(\frac{29.5-20}{4.0825}\right)=0.00998$$

> pnorm(20.5/4.0825)-pnorm(9.5/4.0825)
[1] 0.00998223

and similarly for the other problems.

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  • $\begingroup$ Hello. These values are indeed higher than the ones that I've obtained. They are quite far off actually. I have 0.0071, 0.6879, and 0 respectively. I obtained z-scores of -0.49 and 7.35 for part b and 12.25 for part c respectively. Am I doing something wrong or is the question itself full of bad numbers? What I do is I take the p values of z-scores 7.35 and 12.25 as 1. $\endgroup$ Oct 12, 2018 at 10:53

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