All categories

3 years ago

Round 7,883 to nearest thousand

Mathematics

2 answers:

klasskru [66]3 years ago

7 0

Hi!

Your answer will be: 8,000.

hope this helps!

Stels [109]3 years ago

6 0

Answer:

8,000

Step-by-step explanation:

the nearest thousand is 8,000 as it is past 7,500.

You might be interested in

Syyshshdhehqvswwpwurxwittau

OlgaM077 [116]

Step-by-step explanation:

the picture isn't realistic but mathematically it's:

1) = 121°

2) = 59°

5 0

2 years ago

The time a randomly selected individual waits for an elevator in an office building has a uniform distribution with a mean of 0.

Amiraneli [1.4K]

Answer:

The mean of the sampling distribution of means for SRS of size 50 is $\mu = 0.5$ and the standard deviation is $s = 0.0409$

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size, of at least 30, can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 0.5, \sigma = 0.289$

What are the mean and standard deviation of the sampling distribution of means for SRS of size 50?

By the Central Limit Theorem

$\mu = 0.5, s = \frac{0.289}{\sqrt{50}} = 0.0409$

The mean of the sampling distribution of means for SRS of size 50 is $\mu = 0.5$ and the standard deviation is $s = 0.0409$

Does it matter that the underlying population distribution is not normal?

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

What is the probability a sample of 50 people will wait longer than 45 seconds for an elevator?

We have to use 45 seconds as minutes, since the mean and the standard deviation are in minutes.

Each minute has 60 seconds.

So 45 seconds is 45/60 = 0.75 min.

This probability is 1 subtracted by the pvalue of Z when X = 0.75. So

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

By the Central Limit Theorem

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