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

Question

vampirchik

2 months ago

12

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)

Mathematics

1 answer:

Zina [12.3K] · Answer 1 · 2026-03-10T17:54:16+03:00

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:

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