Question

Suppose you flip a fair coin N times. Let random variable h be the number of heads that occur. Use the normal approximation to estimate the following probabilities. Write your answers using integrals. Do not evaluate the integrals. P(h E [495000, 505000]) given that = 100 b. P(h > 9000) given that N = 104 P(h < 40 or h > 60) given that N = 102 а. C.

Answer 1

Answer:

If n = 1000000, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If n = 10400, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If N = 102, then

$P(h < 40 or h > 60) = 1-P(40$

Step-by-step explanation:

Since the coin is fair, then the probability that a filp is heads is 1/2. Given N tries, the amount of heads can be approximated with a Normal distribution with mean μ =  N *1/2 = N/2 and standard deviation σ = √(N*1/2 * 1/2) = √N/ 2

The density function of that random variable is given by de following formula

$f_X(k) = \frac{1}{\sqrt{2\pi} * \sigma} e^{\frac{-(k-\mu)^2}{2\sigma^2} } = \frac{2}{\sqrt{2\pi N}} e^{\frac{-2(k-N/2)^2}{N} }$

If n = 1000000, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If n = 10400, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If N = 102, then

$P(h < 40 or h > 60) = 1-P(40$