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2 years ago

An exam consists of 50 multiple choice questions. Based on how much you studied, for

1 answer:

5 0

Using the Central Limit Theorem, it is found that the sampling distribution of the sample proportion of the 50 questions on which you get the correct is approximately normal, with mean of 0.7 and standard error of 0.0648.

<h3>What does the Central Limit Theorem state?</h3>

It states that 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, we have that p = 0.7, n = 50, hence the mean and the standard deviation are given as follows:

$\mu = p = 0.7$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.7(0.3)}{50}} = 0.0648$

More can be learned about the Central Limit Theorem at brainly.com/question/24663213

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how many cheese pizzas can the girls buy if they pool all their money together they have $50.00 total and the cheese pizza 7.25

choli [55]

The girls could buy 6 cheese pizzas

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3 years ago

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

What’s the measure of

Wewaii [24]

Answer:

which angle ??

if it is DEC Then it will be 75°

4 0

3 years ago

Read 2 more answers

Describe the relationship between the 6

Maru [420]

The relationship between the 6 in 45,369 and the 6 in 67,420 is that; Clearly, they are not the same values. When written out, 45,369 has a 6 in the tens place, whereas 67,420 has a 6 in the hundreds of thousands place. This is further explained below.

<h3>What are relations b/w numbers?</h3>

Generally, A percentage is another name for this concept. One variant of this word is the "percent," which refers to the ratio of the number of items in a collection to the total number of items in the collection.

In conclusion, natural numbers are a component of the number system that counts from one to infinity and contains all positive integers from one to infinity. Natural numbers are also utilized for the purpose of counting. It does not contain zero (0). In point of fact, the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, etc. are sometimes referred to as counting numbers.

The six that are found in the number 45,369 and the six that are found in the number 67,420 are related in the sense that; Obviously, these are not the same values. In written form, the number 67,420 has a six in the tens place, but the number 45,369 has a six in the hundreds of thousands place.