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
The number that should go in the box is 6.
3.60
- 0.65
---------------
2. 9 5
Y=X/5
Step-by-step explanation:
Step 1:
Let X be the input and Y be the output variable
Given: the output is one-fifth of the input
Step 2:
Y=X/5 be the require function
Eq:
for X=1, Y = 1/5
X=2, Y=2/5, etc
Answer:
the answer is
The point (1, −5) lie in the solution set