According to sources, the most probable answer to this query is that thee results seems to support a .05 probability. Meaning, this supports the hypothesis' claim. Thank you for your question. Please don't hesitate to ask in Brainly your queries.

6 0

2 years ago

What are compatible numbers?

castortr0y [4]

Compatible numbers are numbers that are close in value to the actual numbers and easy to add, subtract, multiply, or divide mentally. They are useful in estimating the sum, difference, product, or quotient.

7 0

3 years ago

The exam scores for 200 students are normally distributed with a mean of 72 and a standard deviation of 10. Which answer choice

Paha777 [63]

Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 72, \sigma = 10$

The proportion of students with scores between 72 and 82 is the <u>p-value of Z when X = 82 subtracted by the p-value of Z when X = 72</u>.

X = 82:

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

$Z = \frac{82 - 72}{10}$

Z = 1

Z = 1 has a p-value of 0.84.

X = 72:

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

$Z = \frac{72 - 72}{10}$

Z = 0

Z = 0 has a p-value of 0.5.

0.84 - 0.5 = 0.34.

Out of 200 students, the number is given by:

0.34 x 200 = 68 students with scores between 72 and 82.

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

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6 0

2 years ago