BODMAS QUESTIONS<br>10-4×0+3÷3

A professor using an open source introductory statistics book predicts that 10% of the students will purchase a hard copy of the

Likurg_2 [28]

Answer: The professor was not accurate with his hypothesis.

Null hypothesis: P1 = 12.5%, P2 = 42.5%, P3 = 45%

The alternate hypothesis: At least one proportion of the student will differ from the others.

Step-by-step explanation: To check if the professors hypothesis were inaccurate.

What percentage of student bought a hard copy of the book.

(25 ÷ 200) × 100 = 12.5%

What percentage of the student printed it from the web.

(85 ÷ 200) × 100 = 42.5%

What percentage of the students read it online.

(90 ÷ 200) × 100 = 45%

This means that the professor was not accurate with his hypothesis. Because the proportion of student in his hypothesis was not the same in the actual.

Therefore; the null hypothesis are

P1 = 12.5%, P2 = 42.5%, P3 = 45%

The alternative hypothesis will state that at least one of the proportion will be different from the others.

5 0

3 years ago

An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in t

Natali5045456 [20]

Answer:

a) Kenji's time had a lower z-score, which means that he is better compared to other runners on his level, and better to his peers compared to Nedda.

b) Rachel, because her time had the lowest z-score.

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

The lower the time, the better the runner is. So whoever's time has a lower z-score is a better runner. So initially, we find the z-score of each student's time.

An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in the class, ran 1 mile in 8 minutes.

This means that for Rachel's time, we have that $X = 8, \mu = 11, \sigma = 3$ . So

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

$Z = \frac{8 - 11}{3}$

$Z = -1$

A junior high school class ran 1 mile in an average of 9 minutes, with a standard deviation of 2 minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes.

This means that for Kenji's time, we have that $X = 8.5, \mu = 9, \sigma = 2$ . So

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

$Z = \frac{8.5 - 9}{2}$

$Z = -0.25$

A high school class ran 1 mile in an average of 7 minutes with a standard deviation of 4 minutes. Nedda, a student in the class, ran 1 mile in 8 minutes.

This means that for Nedda's time, we have that $X = 8, \mu = 7, \sigma = 4$ . So