The average and standard deviation of a measurement is 15.0 ± 5.0. if 100 students measured the value, how many would you expect
1 answer:
Solution: We are given:
Using the empirical rule, we have:
covers 68% of data.
Also the percentage of values below mean = Percentage of values above mean = 50%
Now, let's find the z score for x=20
Therefore, the percentage of values greater than 1 standard deviation above mean
Expected number of students = 16% of 100 = 16
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Let h = amount of hours
Since they earn $7.50 per hour, we can multiply that by an unknown amount of hours to get $116.25. Let's solve:
Divide both sides by 7.5 to isolate the h, amount of hours.
Simplify.
In conclusion, someone would have to work 15.5 hours to earn $116.25 if they earn $7.50 per hour.
Using the normal distribution, it is found that 0.26% of the items will either weigh less than 87 grams or more than 93 grams.
In a <em>normal distribution</em> with mean and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
In this problem:
- The mean is of 90 grams, hence
.
- The standard deviation is of 1 gram, hence
.
We want to find the probability of an item <u>differing more than 3 grams from the mean</u>, hence:
The probability is P(|Z| > 3), which is 2 multiplied by the p-value of Z = -3.
- Looking at the z-table, Z = -3 has a p-value of 0.0013.
2 x 0.0013 = 0.0026
0.0026 x 100% = 0.26%
0.26% of the items will either weigh less than 87 grams or more than 93 grams.
For more on the normal distribution, you can check brainly.com/question/24663213