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Q.105

Expert-verifiedFound in: Page 152

Book edition
OER 2018

Author(s)
Barbara Illowsky, Susan Dean

Pages
902 pages

ISBN
9781938168208

An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes.

a. Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he?

b. Who is the fastest runner with respect to his or her class? Explain why.

a) Since Kenji's time for one mile was 0.25 standard deviations faster than the mean of his class and Nedda's time was 0.25 standard deviations slower than her class, Kenji is regarded a better runner than Nedda.

b) Rachel was the fastest runner in her class, with a time that was one standard deviation faster than the rest of the teams.

Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he.

The mean is one of the measurements of central tendency in statistics. It's calculated by dividing the total number of observations by the sum of the observations. One of the measurements of dispersion is the standard deviation, which is used to show how much individuals in a group differ from the mean.

The z-score can be used to compare a student's performance when different types of scores are being considered.

The Z-score indicates how far a value is above or below the mean in terms of standard deviation. The Z-score is calculated using the following formula:

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

Where:

$x=$ is any data value

$\mu =$ is the population mean

$\sigma =$ is the population standard deviation

In the given example, we are given the below information:

Student | Time (min) | Average time | Standard deviation |

Rachel | $8$ | $11$ | $3$ |

Kenji | $8.5$ | $9$ | $2$ |

Nedda | $8$ | $7$ | $4$ |

Now find the Z- score for the each of the above Students.

The z-score for Rachel is:

$z=\frac{8-11}{3}=-1$

Hence, the Rachel took -1 standard deviation below the mean time of elementary class.

The z-score for Kenji is:

$z=\frac{8.5-9}{2}=-0.25$

Therefore, the Kenji took - 0.25 standard deviations below the mean time of junior class.

The z-score for Nedda is:

$z=\frac{8-7}{4}=0.25$

Hence, the Nedda took 0.25 standard deviations above the mean time of high school class.

As Kenji's time for one mile was 0.25 standard deviations faster than the mean of his class and Nedda's time was 0.25 standard deviations slower than her class, we can clearly see that Kenji is a better runner than Nedda.

Who is the fastest runner with respect to his or her class.

The mean is one of the measurements of central tendency in statistics. It's calculated by dividing the total number of observations by the sum of the observations. One of the measurements of dispersion is the standard deviation, which is used to show how much individuals in a group differ from the mean.

The z-score can be used to compare a student's performance when different types of scores are being considered.

The Z-score indicates how far a value is above or below the mean in terms of standard deviation. The Z-score is calculated using the following formula:

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

Where

$x=$ is any data value

$\mu =$ is the population mean

$\sigma =$ is the population standard deviation

In the given example, we are given the below information:

Student | Time(mins) | Average time | Standard deviation |

Rachel | $8$ | $11$ | $3$ |

Kenji | $8.5$ | $9$ | $2$ |

Nedda | $8$ | $7$ | $4$ |

Now find the Z- score for the each of the above Students.

The z-score for Rachel is:

$z=\frac{8-11}{3}=-1$

Hence, the Rachel took -1 standard deviation below the mean time of elementary class.

The z-score for Kenji is:

$z=\frac{8.5-9}{2}=-0.25$

Therefore, the Kenji took - 0.25 standard deviations below the mean time of junior class.

The z-score for Nedda is:

$z=\frac{8-7}{4}=0.25$

Hence, the Nedda took 0.25 standard deviations above the mean time of high school class.

We can see from above z-score numbers that Rachel had the lowest z-score of the three, -1, indicating that she had a time that was one standard deviation faster than her class. As a result, Rachel was the quickest runner in her class out of the three students.

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