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2 years ago

Can someone help me with this one please

Mathematics

1 answer:

Vera_Pavlovna [14]2 years ago

6 0

Im not sure what it means by the entire triangular region so ill tell you that the highlighted right triangle's area is 60, and the area for both triangles is 120. semicircle area is 201.06. total area for entire shape is 321.06.

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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.

8 0

3 years ago

Q1. Ten percent of the items produced by a machine are defective. A random sample of 100 items is selected and checked for defec

Tcecarenko [31]

Answer: 0.030

Step-by-step explanation:

Given: The percent of the items produced by a machine are defective : $P=10\%=0.10$

The percent of the items produced by machine which are not defective: $Q=1-0.1=0.9$

Sample size : n = 100

Now, the standard error of the proportion is given by :-

$\text{S.E.}=\sqrt{\dfrac{0.1\times0.9}{100}}\\\\\Rightarrow\text{S.E.}=0.030$

Hence, the the standard error of the distribution of the sample proportions=0.030

7 0

2 years ago

Find the slope-intercept form of the line that passes through given the point and has the given slope. Include your work in your

nignag [31]

Slope-intercept form is y=mx+b
we know m is 10 and we have x and y
if you plug in x, y and m and solve for b, like this
y=mx+b
1=10 (7) + b
1=70+b
-69=b
so now we know b is -69. just plug that in
the final answer is y=10x-69

6 0

2 years ago

A book store is keeping track of the price of books according to the number of pages. Which scatterplot represents that bivariat

tangare [24]

Answer:

The graph has pages on the X-axis and price (dollars) on the Y-axis

Step-by-step explanation:

The X-axis will always have the independent variable (pages), while the Y-axis will have the dependent variable (number of pages). The pages will exist no matter how many there are, but the price is influenced by the number of pages.

3 0

2 years ago

Read 2 more answers

1. An AP Statistics class starts a project to estimate the average number of hours a student at their high school sleeps per nig

Vladimir [108]

Answer:

Option E

Step-by-step explanation:

This survey will not be reliable as it is chosen for convenience; the first 120 students to arrive in school on a particular morning and this collection of individuals may not be a representative of the population. The study may become biased because it does not take into account the latecomers among the students which might have been changed the study from systematically favoring certain outcomes.

5 0

3 years ago