Answer:
2, 4, 5
Step-by-step explanation:
Answer:
12
Step-by-step explanation:
A score of 85 would be 1 standard deviation from the mean, 74. Using the 68-95-99.7 rule, we know that 68% of normally distributed data falls within 1 standard deviation of the mean. This means that 100%-68% = 32% of the data is either higher or lower. 32/2 = 16% of the data will be higher than 1 standard deviation from the mean and 16% of the data will be lower than 1 standard deviation from the mean. This means that 16% of the graduating seniors should have a score above 85%.
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