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17 days ago

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John averages 58 words per minute on a typing test with a standard deviation of 11 words per minute. Suppose John's words per mi

nute on a minute on a typing test. Then X~N(58,11)

1 answer:

Zina [9.1K]17 days ago

8 0

Answer:

When x = 72, the z score is:

$z = \frac{72-58}{11}= 1.273$

The average is 58

This z-score indicates that x= 72 is 1.273 standard deviations above the mean.

Step-by-step explanation:

Let’s consider the following details for this question: John's typing speed on a test is assumed to follow a normal distribution. Let X denote the number of words he can type in a minute. Hence, X ~ N(58, 11). It’s important to round the result to three decimal points if needed.

Provide your response below for words per minute on a typing test conducted on Sunday. The z score when x = 72 is

In this scenario, we acknowledge that the variable under consideration is represented by a normal distribution:

$X \sim N (\mu= 58, \sigma=11)$

The formula for the z score is as follows:

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

By substituting values, we derive:

$z = \frac{72-58}{11}= 1.273$

The mean is 58

This z-score indicates that x= 72 is 1.273 standard deviations higher than the mean.

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$\frac{2xx_0}{a^2}+\frac{2yy_0}{b^2}-\frac{2zz_0}{c^2}-2\left(\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}-\frac{z_0^2}{c^2}\right)=0$

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Explicación paso a paso:

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Read 2 more answers