Respuesta :
Answer:
(a) The probability that proportion of heads is between 0.30 and 0.70 is 1.
(b) The probability that proportion of heads is between 0.40 and 0.65 is 0.9759.
Step-by-step explanation:
Let X = number of heads.
The probability that a head occurs in a toss of a coin is, p = 0.50.
The coin was tossed n = 100 times.
A random toss's result is independent of the other tosses.
The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.50.
But the sample selected is too large and the probability of success is 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of [tex]\hat p[/tex] (sample proportion of X) if the following conditions are satisfied:
- np ≥ 10
- n(1 - p) ≥ 10
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]\hat p\sim N(p,\ \frac{p(1-p)}{n})[/tex]
[tex]\mu_{p}=p=0.50\\\sigma_{p}=\sqrt{\frac{p(1-p)}{n}}=0.05[/tex]
(a)
Compute the probability that proportion of heads is between 0.30 and 0.70 as follows:
[tex]P(0.30<\hat p<0.70)=P(\frac{0.30-0.50}{0.05}<\frac{\hat p-p}{\sigma_{p}}<\frac{0.70-0.50}{0.05})\\[/tex]
[tex]=P(-4<Z<4)\\=P(Z<4)-P(Z<-4)\\=(\approx1)-(\approx0)\\=1[/tex]
Thus, the probability that proportion of heads is between 0.30 and 0.70 is 1.
(b)
Compute the probability that proportion of heads is between 0.40 and 0.65 as follows:
[tex]P(0.40<\hat p<0.65)=P(\frac{0.40-0.50}{0.05}<\frac{\hat p-p}{\sigma_{p}}<\frac{0.65-0.50}{0.05})\\[/tex]
[tex]=P(-2<Z<3)\\=P(Z<3)-P(Z<-2)\\=0.9987-0.0228\\=0.9759[/tex]
Thus, the probability that proportion of heads is between 0.40 and 0.65 is 0.9759.
