Question

6.8 Use the Normal approximation. Suppose we toss a fair coin 100 times. Use the Normal approximation to find the probability that the sample proportion of heads is (a) between 0.3 and 0.7. (b) between 0.4 and 0.65. Moore, David. Exploring the Practice of Statistics & Student CD (p. 325). W.H. Freeman & Company. Kindle Edition.

Answer 1

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 $\hat p$ (sample proportion of X) if the following conditions are satisfied:

1. np ≥ 10
2. n(1 - p) ≥ 10

Check the conditions as follows:

 $np=100\times 0.50=50>10\\n(1-p)=100\times (1-0.50)=50>10$

Thus, a Normal approximation to binomial can be applied.

So,  $\hat p\sim N(p,\ \frac{p(1-p)}{n})$

$\mu_{p}=p=0.50\\\sigma_{p}=\sqrt{\frac{p(1-p)}{n}}=0.05$

(a)

Compute the probability that proportion of heads is between 0.30 and 0.70 as follows:

$P(0.30$

                              $=P(-4$

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:

$P(0.40$

                              $=P(-2$

Thus, the probability that proportion of heads is between 0.40 and 0.65 is 0.9759.