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

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

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* 6x(1-x)} \, dx \\\\=\int\limits^(1)_(0) {6x^(2)-6x^(3)} \, dx\\\\=[(6x^(3))/(3)]-[(6x^(4))/(4)]|^(1)_(0)\\\\=2-(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)* 6x(1-x)} \, dx \\\\=\int\limits^(1)_(0) {6x^(3)-6x^(4)} \, dx\\\\=[(6x^(4))/(4)]-[(6x^(5))/(5)]|^(1)_(0)\\\\=(3)/(2)-(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)=(\sigma)/(√(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((0.45-0.50)/(0.2236/√(64))<(\bar X-\mu_(\bar x))/(\sigma_(\bar x))<(0.50-0.50)/(0.2236/√(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.