Answer:
There is a 0.82% probability that a line width is greater than 0.62 micrometer.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by
After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.
In this problem
The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so 
.
What is the probability that a line width is greater than 0.62 micrometer?
That is 
So
Z = 2.4 has a pvalue of 0.99180.
This means that P(X \leq 0.62) = 0.99180.
We also have that
There is a 0.82% probability that a line width is greater than 0.62 micrometer.
I can’t see your question I wish I could help you
The random sample of the students is an illustration of sampling
The chi-square test for goodness of fit is inappropriate because the variable under study is not categorical.
<h3>How to determine the reason chi square is not appropriate?</h3>
The dataset is given as:
Monday 34
Tuesday 29
Wednesday 32
Thursday 28
Friday 19
The variable of the above dataset is a not a categorical dataset.
One of the conditions of the chi-square test for goodness of fit test is that the variable under study must be categorical.
Hence, the chi-square test for goodness of fit is inappropriate because the variable under study is not categorical.
Read more about chi-square test at:
brainly.com/question/19959558
Answer:
the answer is 3 feet
Step-by-step explanation:
We can use the volume formula, V =1/2 bhl to find the height of the tent
the volume is 
. The base is 4 and the length is 6 feet
36=1/2 ⋅4⋅h⋅6
36=12h
3=h
The height of <em>the tent is 3 feet.</em>
