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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
Answer:
Step-by-step explanation:
Given equation can be re written as
............(i)
Now it is given that y(π/2) = 2e
Applying value in (i) we get
ln(2e) = sin(π/2) + c
=> ln(2) + ln(e) = 1+c
=> ln(2) + 1 = 1 + c
=> c = ln(2)
Thus equation (i) becomes
ln(y) = sin(x) + ln(2)
ln(y) - ln(2) = sin(x)
ln(y/2) = sin(x)