The weights of certain machine components are normally distributed with a mean of 8.5 g and a standard deviation of 0.09 g. Find
1 answer:
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Answer:
n = 2
Step-by-step explanation:
In order for the ratio to have a value of 1, the sum of exponents in the numerator must equal the sum of exponents in the denominator:
n + 7 = 9
n = 9 -7 . . . . . . subtract 7 from both sides of the equation
n = 2
___
Alternatively, you can simplify the given expression:
Hopefully, it is clear to you that the value of n must be 2 in order to make the numerator equal to the denominator.
Answer:
a) 18.94% probability that the sample mean amount purchased is at least 12 gallons
b) 81.06% probability that the total amount of gasoline purchased is at most 600 gallons.
c) The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.
Step-by-step explanation:
To solve this question, we use the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, 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.
For sums, we can apply the theorem, with mean and standard deviation
In this problem, we have that:
a. In a sample of 50 randomly selected customers, what is the approximate probability that the sample mean amount purchased is at least 12 gallons?
Here we have
This probability is 1 subtracted by the pvalue of Z when X = 12.
By the Central Limit theorem
has a pvalue of 0.8106.
1 - 0.8106 = 0.1894
18.94% probability that the sample mean amount purchased is at least 12 gallons
b. In a sample of 50 randomly selected customers, what is the approximate probability that the total amount of gasoline purchased is at most 600 gallons.
For sums, so
This probability is the pvalue of Z when X = 600. So
has a pvalue of 0.8106.
81.06% probability that the total amount of gasoline purchased is at most 600 gallons.
c. What is the approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers.
This is X when Z has a pvalue of 0.95. So it is X when Z = 1.645.
The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.