Question

It is known that 70% of all brand A external hard drives work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that n = 15 drives are randomly selected. Let X = the number of successes in the sample. The statistic X/n is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. [Hint: One possible value X/n is 0.2, corresponding to X = 3. What is the probability of this value (what kind of random variable is X)?] (Round your answers to three decimal places.)

Answer 1

Answer:

The probability distribution of $\hat p$ is, $N(0.70, 0.032)$.

Step-by-step explanation:

The random variable X is defined as the number of brand A external hard drives work in a satisfactory manner throughout the warranty period.

The sample selected is of size, n = 15.

The probability of selecting a hard drive that works in a satisfactory manner throughout is, p = 0.70.

Every hard drive works independently of the others.

The random variable X follows a Binomial distribution with parameters n = 15 and p = 0.70.

The probability mass function of X, the binomial random variable is:

$P(X=x)={15\choose x}0.70^{x}(1-0.70)^{15-x};\ x=0,1,2,3...$

The probability distribution of X is shown below.

Now a statistic is defined as:

$\hat p=\frac{X}{n}$

This statistic is known as the sample proportion.

A Normal approximation to Binomial can be used to approximate the distribution of sample proportion if the following conditions are satisfied:

• np ≥ 5
• n(1 - p) ≥ 5

Then the sampling distribution of $\hat p$ follows a Normal; distribution with mean $p=0.70$ and standard deviation $\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.70(1-0.70)}{200}}=0.032$.

Check the conditions:

$np=15\times0.70=10.5\\n(1-p)=15\times (1-0.70)=4.5\approx5$

So, the probability distribution of $\hat p$ is, $N(0.70, 0.032)$.