Answer:
96.42% probability that the number of tails is between 40 and 60.
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that
,
.
In this problem, we have that:
100 tosses, so ![n = 100](https://tex.z-dn.net/?f=n%20%3D%20100)
Two outcomes, both equally as likely. So ![p = \frac{1}{2} = 0.5](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%3D%200.5)
So
![E(X) = np = 100*0.5 = 50](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np%20%3D%20100%2A0.5%20%3D%2050)
![\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.5*0.5} = 5](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B100%2A0.5%2A0.5%7D%20%3D%205)
Estimate the approximate probability that the number of tails is between 40 and 60.
Using continuity correction.
![P(40 - 0.5 \leq X \leq 60 + 0.5) = P(39.5 \leq X \leq 60.5)](https://tex.z-dn.net/?f=P%2840%20-%200.5%20%5Cleq%20X%20%5Cleq%2060%20%2B%200.5%29%20%3D%20P%2839.5%20%5Cleq%20X%20%5Cleq%2060.5%29)
This is the pvalue of Z when X = 60.5 subtracted by the pvalue of Z when X = 39.5. So
X = 60.5
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{60.5 - 50}{5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B60.5%20-%2050%7D%7B5%7D)
![Z = 2.1](https://tex.z-dn.net/?f=Z%20%3D%202.1)
has a pvalue of 0.9821
X = 39.5
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{39.5 - 50}{5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B39.5%20-%2050%7D%7B5%7D)
![Z = -2.1](https://tex.z-dn.net/?f=Z%20%3D%20-2.1)
has a pvalue of 0.0179
0.9821 - 0.0179 = 0.9642
96.42% probability that the number of tails is between 40 and 60.