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FinnZ [79.3K]
3 years ago
11

I need help on number 4

Mathematics
1 answer:
podryga [215]3 years ago
4 0
The final answer/value of x is 6.

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Let X represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the me
Alexus [3.1K]

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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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 \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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 \mu and standard deviation s = \sqrt{n}*\sigma

In this problem, we have that:

\mu = 11.5, \sigma = 4

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 n = 50, s = \frac{4}{\sqrt{50}} = 0.5657

This probability is 1 subtracted by the pvalue of Z when X = 12.

Z = \frac{X - \mu}{\sigma}

By the Central Limit theorem

Z = \frac{X - \mu}{s}

Z = \frac{12 - 11.5}{0.5657}

Z = 0.88

Z = 0.88 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 mu = 50*11.5 = 575, s = \sqrt{50}*4 = 28.28

This probability is the pvalue of Z when X = 600. So

Z = \frac{X - \mu}{s}

Z = \frac{600 - 575}{28.28}

Z = 0.88

Z = 0.88 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.

Z = \frac{X - \mu}{s}

1.645 = \frac{X- 575}{28.28}

X - 575 = 28.28*1.645

X = 621.5

The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

5 0
3 years ago
Simplify the square root of 9 ÷ by the square root of 25
Zanzabum
So pemdas 
square root of 9 means 9^(1/2) so it counts as an exponent so 
9=3 times 3 so square root of 9=3
square root of 25
do pemdas so exponents before division
25=5 times 5 so 
square root of 25==5 so 

3/5=0.6
5 0
3 years ago
Read 3 more answers
Find the equation of the line whose slope is 1/3
Masja [62]

Answer:

It means that when x = 3, y = 1, obviously when the first function expression takes 3, y = 1

3 0
3 years ago
Ahh help, I need this question due in 10 mins-
sp2606 [1]
I believe its 838 inches squared
8 0
3 years ago
I need help please plzzzzzzzzzzzzzzzzzz
ladessa [460]

Answer:

One solution, 8x + 9

Step-by-step explanation:

3x + 9 + 4x + x

Add the x's

8x + 9

<em>good luck, hope this helps :)</em>

8 0
3 years ago
Read 2 more answers
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