What is the probability that in 120 tosses of a fair coin between 40% and 60% (inclusively) will be heads?
Note,Originally Posted by swoopesjr01
40% of 120 is 48
60% of 120 is 72
Which means the probability is,
$\displaystyle P(48)+P(49)+...+P(71)+P(72)=\sum^{72}_{k=48}P(k)$
But, because of binomial probability we have,
$\displaystyle P(k)={120 \choose k}(1/2)^k(1/2)^{120-k}={120 \choose k}(1/2)^{120}$
Thus,
$\displaystyle \sum^{72}_{k=48}{120 \choose k}(1/2)^{120}\approx .9779$
Use the normal approximation to the Binomial distribution for this.Originally Posted by swoopesjr01
The mean number of heads is $\displaystyle 60$, the standard deviation is
$\displaystyle \sqrt{120\times 0.5\times 0.5}=\sqrt{30}$
Between 40% and 60% of 120 is between 48 and 72. To allow for the
fact that we can only have an integer number of heads we work the
normal approximation (which is continuous) for between 47.5 and 72,5
heads.
The z-scores for the ends of the range are:
$\displaystyle
z_{lo}=\frac{47.5-60}{\sqrt{30}}\approx -2.282
$
$\displaystyle
z_{hi}=\frac{72.5-60}{\sqrt{30}}\approx 2.282
$
Now looking the probability of a standard normal random variable having
a value in the range $\displaystyle [-2.28,2.28]$ we find the probability is
$\displaystyle \approx 0.9775$.
This last answer may be compared with PH's calculation based on using
the exact binomial formula.
RonL