What can 385/64 be simplified to?

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and

masya89 [10]

Answer:

a) There is a 100% probability that the (sample) average time waiting in line for these customers is less than 10 minutes.

b) There is a 100% probability that the (sample) average time waiting in line for these customers is between 5 and 10 minutes.

c) There is a 0% probability that the (sample) average time waiting in line for these customers is less than 6 minutes.

d) Because there are less observations, it would be less accurate.

e) Because there are moreobservations, it would be more accurate.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. This means that $\mu = 8.2, \sigma = 1.5$ .

Suppose that a random sample of n = 49 customers is observed

This means that $s = \frac{1.5}{\sqrt{49}} = 0.21$ .

(a) Less than 10 minutes.

This probability is the pvalue of Z when $X = 10$ . So:

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

$Z = \frac{10 - 8.2}{0.21}$

$Z = 8.57$

$Z = 8.57$ has a pvalue of 1.

This means that there is a 100% probability that the (sample) average time waiting in line for these customers is less than 10 minutes.

(b) Between 5 and 10 minutes.

This probability is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 5.

From a), we have that the zscore of X = 10 has a pvalue of 1.

For X = 5.

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

$Z = \frac{5 - 8.2}{0.21}$

$Z = -15.24$

$Z = -15.24$ has a pvalue of 0.

Subtracting, we have that there is a 100% probability that the (sample) average time waiting in line for these customers is between 5 and 10 minutes.

(c) Less than 6 minutes.

This probability is the pvalue of Z when X = 6. So:

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

$Z = \frac{6 - 8.2}{0.21}$

$Z = -10.48$

$Z = -10.48$ has a pvalue of 0.

This means that there is a 0% probability that the (sample) average time waiting in line for these customers is less than 6 minutes.

(d) If you only had two observations instead of 49 observations, would you believe that your answers to parts (a), (b), and (c), are more accurate or less accurate? Why?

The less observations there are, the less acurrate our results are.

So, because there are less observations, it would be less accurate.

(e) If you had 1,000 observations instead of 49 observations, would you believe that your answers to parts (a), (b), and (c), are more accurate or less accurate? Why?

The more observations there are, the more acurrate our results are.

So, because there are moreobservations, it would be more accurate.

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