Respuesta :
Answer:
(a) 0.1585
(b) 0.4713
(c) 0.9545
Explanation:
The random variable X can be defined as the number of heads.
The coin provided is balanced, i.e. P (H) = P (T) = 0.50
The outcome of tossing the coin are: (H and T). Each of these outcomes are independent of each other.
The random variable X thus follows a Binomial distribution with probability of success as 0.50.
For a large sample a Normal approximation to binomial can be applied to approximate the distribution of p if the following conditions are satisfied:
1. np ≥ 10
2. n(1 - p) ≥ 10
(a)
n = 100
Check the conditions as follows:
[tex]np=100\times 0.50=50>10\\\\n(1-p)=100\times(1-0.50)=50>10[/tex]
Thus, a Normal approximation to binomial can be applied.
So, [tex]p\sim N(0.50,\ 0.05 )[/tex]
Compute the probability hat the proportion of heads will be anywhere from 0.49 to 0.51 as follows:
[tex]P(0.49<p<0.51)=P(\frac{0.49-0.50}{0.05}<\frac{p-\mu}{\sigma}<\frac{0.51-0.50}{0.05})[/tex]
[tex]=P(-0.20<Z<0.20)\\\\=P(Z<0.20)-P(Z<-0.20)\\\\=0.57926-0.42074\\\\=0.15852\\\\\approx 0.1585[/tex]
Thus, the probability hat the proportion of heads will be anywhere from 0.49 to 0.51 when a balanced coin is flipped 100 times is 0.1585.
(b)
n = 1000
Check the conditions as follows:
[tex]np=1000\times 0.50=500>10\\\\n(1-p)=1000\times(1-0.50)=500>10[/tex]
Thus, a Normal approximation to binomial can be applied.
So, [tex]p\sim N(0.50,\ 0.016 )[/tex]
Compute the probability hat the proportion of heads will be anywhere from 0.49 to 0.51 as follows:
[tex]P(0.49<p<0.51)=P(\frac{0.49-0.50}{0.016}<\frac{p-\mu}{\sigma}<\frac{0.51-0.50}{0.016})[/tex]
[tex]=P(-0.63<Z<0.63)\\\\=P(Z<0.63)-P(Z<-0.63)\\\\=0.73565-0.26435\\\\=0.4713[/tex]
Thus, the probability hat the proportion of heads will be anywhere from 0.49 to 0.51 when a balanced coin is flipped 1000 times is 0.4713.
(c)
n = 10,000
Check the conditions as follows:
[tex]np=10000\times 0.50=5000>10\\\\n(1-p)=10000\times(1-0.50)=5000>10[/tex]
Thus, a Normal approximation to binomial can be applied.
So, [tex]p\sim N(0.50,\ 0.005)[/tex]
Compute the probability hat the proportion of heads will be anywhere from 0.49 to 0.51 as follows:
[tex]P(0.49<p<0.51)=P(\frac{0.49-0.50}{0.005}<\frac{p-\mu}{\sigma}<\frac{0.51-0.50}{0.005})[/tex]
[tex]=P(-2<Z<2)\\\\=P(Z<2)-P(Z<-2)\\\\=0.97725-0.02275\\\\=0.9545[/tex]
Thus, the probability hat the proportion of heads will be anywhere from 0.49 to 0.51 when a balanced coin is flipped 10,000 times is 0.9545.
