Question

A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the true proportion is 0.04. If 469 are sampled, what is the probability that the sample proportion will be less than 0.03

Answer 1

Answer:

Probability that the sample proportion will be less than 0.03 is 0.10204.

Step-by-step explanation:

We are given that a courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the true proportion is 0.04.

Also, 469 are sampled.

Let $\hat p$ = sample proportion

The z-score probability distribution for sample proportion is given by;

              Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion

           p = true proportion = 0.04

           n = sample size = 469

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the sample proportion will be less than 0.03 is given by = P( $\hat p$ < 0.03)

       P( $\hat p$ < 0.03) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\frac{0.03-0.04}{\sqrt{\frac{0.03(1-0.03)}{469} } }$ ) = P(Z < -1.27) = 1 - P(Z $\leq$ 1.27)

                                                                      = 1 - 0.89796 = 0.10204

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.27 in the z table which has an area of 0.89796.

Therefore, probability that the sample proportion will be less than 0.03 is 0.10204.

Answer 2

Using the normal distribution and the central limit theorem, it is found that there is a 0.1335 = 13.35% probability that the sample proportion will be less than 0.03.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$, as long as $np \geq 10$ and $n(1 - p) \geq 10$.

In this problem:

• The true proportion is of p = 0.04.

• 469 people are sampled, hence n = 469.

The mean and the standard error are given by:

$\mu = p = 0.04$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.04(0.96)}{469}} = 0.009$

The probability that the sample proportion will be less than 0.03 is the p-value of Z when X = 0.03, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.03 - 0.04}{0.009}$

$Z = -1.11$

$Z = -1.11$ has a p-value of 0.1335.

0.1335 = 13.35% probability that the sample proportion will be less than 0.03.

To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213