Find the point on the unit circle that corresponds to 5pi/6

A ski lift is designed with a total load limit of 20,000 pounds. It claims a capacity of 100 persons. An expert in ski lifts thi

Yanka [14]

Answer:

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sum of n values of a distribution, the mean is $\mu \times n$ and the standard deviation is $\sigma\sqrt{n}$

An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.

This means that $\mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300$

If the expert is right, what is the probability that a random sample of 100 independent persons will cause an overload

Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So

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

$Z = \frac{20000 - 20000}{300}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

1 - 0.5 = 0.5

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

5 0

3 years ago

4.48 Same observation, difference sample size: Suppose you conduct a hypothesis test based on a sample where the sample size is

Lemur [1.5K]

Answer:

P-value is lesser in the case when n = 500.

Step-by-step explanation:

The formula for z-test statistic can be written as

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } } =\frac{(x-\mu)\sqrt{n}}{\sigma}$

here, μ = mean

σ= standard deviation, n= sample size, x= variable.

From the relation we can clearly observe that n is directly proportional to test statistic. Thus, as the value of n increases the corresponding test statistic value also increases.

We can also observe that as the test statistic's numerical value increases it is more likely to go into rejection region or in other words its P-value decreases.

Now, for first case when our n is 50 we will have a relatively low chance of accurately representing the population compared to the case when n= 500. Therefore, the P-value will be lesser in the case when n = 500.

6 0

3 years ago

gavin has 460 baseball players in his collection of baseball cards and 15% of the players are pitchers.how many pitchers are in

Kruka [31]

69 is the correct answer

460 divided by 100 times 15 = 69

8 0

3 years ago

14 POINTS + Brainliest to first correct answer with explanation and diagram.

zhuklara [117]

So for example we have

1/2 divided by 1/6

What we will do it’s

Leave the first one As it is

Turn the second fraction (the one you want to divide by) upside down

The change the divide to multiply

Multiply the first fraction by that reciprocal

Simplify the fraction (if needed)

So it’s will be like this

1/2 x 6/1 = 6/2

Simplify it

6/2 =3

I hope it’s will help u ✨

5 0

3 years ago

Julie and Jan’s brother also withdrew money from his savings account on Monday. He made three withdrawals and withdrew $10 more

Zepler [3.9K]

Answer:

possible amounts could be $35, $55, $75

hope this helps!

Step-by-step explanation:

8 0

3 years ago