4(3xy^4)^3/(2x^3y^5)^4
1 answer:
You might be interested in
Answer:
0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Normally distributed with a mean of 118 centimeters and a standard deviation of 8 centimeters.
This means that
Sample of 16 shells
This means that
What is the probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly?"
This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 116.
X = 120
By the Central Limit Theorem
has a pvalue of 0.841
X = 116
has a pvalue of 0.159
0.841 - 0.159 = 0.682
0.682 = 68.2% probability that the average length of these 16 shells will be between 116 and 120 centimeters when the machine is operating "properly".