1.Round off 378811 to the nearest hundreds,thousands and ten thousand

2 answers:

8 0

Snowcat [4.5K]3 years ago

6 0

Given:

The numbers are 378811, 267988, 250260, 196596, 193171.

To find:

The approximate value of the given numbers to the nearest hundreds, thousands and ten thousand.

Solution:

The required values shown in the below table:

Numbers Nearest hundreds Nearest thousands Nearest ten thousand

378811 378800 379000 380000

267988 268000 268000 270000

250260 250300 250000 250000

196596 196600 197000 200000

193171 193200 193000 190000

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Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .