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:50%
Step-by-step explanation: It doesn't matter how many questions there are because for each question, you have a 50% chance of getting it right. So, the probability is 50%.
Answer:
we cant see the data
Step-by-step explanation:
B & c
they both go up while a goes down
Answer:
y'=1/x
Step-by-step explanation:
![d/dx[ ln(u) ]=\frac{1}{u} *du\\y'=\frac{1}{cx} *c\\y'=\frac{1}{x}](https://tex.z-dn.net/?f=d%2Fdx%5B%20ln%28u%29%20%5D%3D%5Cfrac%7B1%7D%7Bu%7D%20%2Adu%5C%5Cy%27%3D%5Cfrac%7B1%7D%7Bcx%7D%20%2Ac%5C%5Cy%27%3D%5Cfrac%7B1%7D%7Bx%7D)