Question

Let X1, X1, ..., X64 be a i.i.d. random sample of size 64 from a distribution with probability density function: f(x) = 6x(1 − x), 0 < x < 1 Approximate P(0.45 < X <¯ 0.5). Hint: find µ and σ 2

Answer 1

Answer:

The approximate value of $P(0.45<\bar X<0.50)$ is 0.3686.

Step-by-step explanation:

The pdf of X is:

$f(x) = 6x(1-x), 0 < x < 1$

Compute the mean as follows:

$\mu=\int\limits^{1}_{0} {x\times 6x(1-x)} \, dx \\\\=\int\limits^{1}_{0} {6x^{2}-6x^{3}} \, dx\\\\=[\frac{6x^{3}}{3}]-[\frac{6x^{4}}{4}]|^{1}_{0}\\\\=2-\frac{3}{2}\\\\=0.50$

Compute the variance as follows:

$\sigma^{2}=E(X^{2})-(\mu)^{2}$

$E(X^{2})=\int\limits^{1}_{0} {x^{2}\times 6x(1-x)} \, dx \\\\=\int\limits^{1}_{0} {6x^{3}-6x^{4}} \, dx\\\\=[\frac{6x^{4}}{4}]-[\frac{6x^{5}}{5}]|^{1}_{0}\\\\=\frac{3}{2}-\frac{6}{5}\\\\=0.30$

$\sigma^{2}=E(X^{2})-(\mu)^{2}\\\\\sigma=\sqrt{E(X^{2})-(\mu)^{2}}\\\\=\sqrt{0.30-(0.50)^{2}}\\\\=0.2236$

According to the Central Limit Theorem if an unknown population is selected with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from this population with replacement, then the distribution of the sample means will be approximately normally.  

Then, the mean of the sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The sample selected is of size n = 64 > 30. Thus a central limit theorem can be applied to approximate the sampling distribution of sample mean.

Compute the value of $P(0.45<\bar X<0.50)$ as follows:

$P(0.45<\bar X<0.50)=P(\frac{0.45-0.50}{0.2236/\sqrt{64}}<\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}<\frac{0.50-0.50}{0.2236/\sqrt{64}})$

                                $=P(-1.12<Z<0)\\\\=P(Z<0)-P(Z<-1.12)\\\\=0.50-0.13136\\\\=0.36864\\\\\approx 0.3686$

Thus, the approximate value of $P(0.45<\bar X<0.50)$ is 0.3686.