Question

Hugo averages 63 words per minute on a typing test with a standard deviation of 8 words per minute. Suppose Hugo's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then, X∼N(63,8).

Suppose Hugo types 79 words per minute in a typing test on Wednesday. The z-score when x=79 is ________. This z-score tells you that x=79 is ________ standard deviations to the ________ (right/left) of the mean, ________.
Correctly fill in the blanks in the statement above.

Answer 1

Main Answer:

The z-score when x=79 is 2.0. This z-score tells you that x=79 is 2.0 standard deviations to the right of the mean, 63.

Step-by-step explanation:

In statistics, the z-score is a measure of how many standard deviations a data point is from the mean of a normal distribution. The formula for calculating the z-score is
`(X - μ) / σ`, where X is the value, μ is the mean, and σ is the standard deviation. In this case, we're given that Hugo's typing speed, X, follows a normal distribution with a mean (μ) of 63 words per minute and a standard deviation (σ) of 8 words per minute.

To find the z-score when Hugo types 79 words per minute, we use the formula:
`\( Z = \frac{{X - \mu}}{{\sigma}} \).` Substituting in the given values, we get
`\( Z = \frac{{79 - 63}}{{8}} = 2.0 \).` This positive value indicates that 79 words per minute is 2.0 standard deviations above the mean.

Interpreting the z-score, we say that Hugo's typing speed of 79 words per minute is 2.0 standard deviations to the right of the mean (63). In a normal distribution, positive z-scores represent values above the mean. Therefore, 79 words per minute is higher than the average typing speed of 63 words per minute.