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bonufazy [111]
3 years ago
14

Estimate the sum by rounding each number to the given place value.

Mathematics
1 answer:
romanna [79]3 years ago
6 0

Answer:

400,000

Step-by-step explanation:

The given sum is 164,215+216,088

We simplify this to get:

164,215+216,088=380,303

The digit at the hundred thousand position is the rightmost 3.

The digit after it is 8.

Since 8≥5 we round up.

We add 1 to 3 to get 4 and replace all digits to the right with zeros.

We round to the nearest hundred thousand to get:

400,000

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Using the normal distribution, it is found that:

1. His z-score was of Z = -1.88.

2. There is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

3. Z-score of z = 1.85, there is a 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

4. Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.

In this problem:

  • The mean is of \mu = 1.35.
  • The standard deviation is of \sigma = 0.33.

Item 1:

Considering his ratio, we have that X = 0.73, hence:

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

Z = \frac{0.73 - 1.35}{0.33}

Z = -1.88

His z-score was of Z = -1.88.

Item 2:

The probability is the <u>p-value of Z = -1.88</u>, hence, there is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

Item 3:

Score of X = 1.96, hence:

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

Z = \frac{1.96 - 1.35}{0.33}

Z = 1.85

The probability is 1 subtracted by the p-value of Z = 1.85, hence, 1 - 0.9678 = 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

Item 4:

Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

More can be learned about the normal distribution at brainly.com/question/24663213

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