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

Question

17 days ago

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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

an value and standard deviation of X are 11.5 and 4.0, respectively.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?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.c. What is the approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers.

Mathematics

1 answer:

babunello [8.4K] · Answer 1 · 2026-03-10T14:08:28+03:00

Answer:

a) There is an 18.94% chance that the sample mean of the amount purchased will be at least 12 gallons.

b) There is an 81.06% chance that the total gasoline purchased will not exceed 600 gallons.

c) The estimated value for the 95th percentile of the total consumption by 50 randomly chosen customers is 621.5 gallons.

Step-by-step explanation:

The solution to this query involves applying the normal probability distribution and the central limit theorem.

Normal probability distribution

Issues involving normally distributed samples can be addressed using the z-score formula.

In a dataset characterized by mean $\mu$ and standard deviation $\sigma$ , the z-score for a value X is expressed as:

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

The z-score indicates how many standard deviations a particular value is from the mean. After calculating the z-score, we reference the z-score table to find its corresponding p-value, which represents the probability that a measure is less than X, essentially giving us X's percentile. By subtracting the p-value from 1, we find the chance that the measure exceeds X.

Central Limit Theorem

The Central Limit Theorem posits that for a normally distributed variable X, with mean $\mu$ and standard deviation $\sigma$ , the distribution of sample means with size n approximates a normal distribution characterized by mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

Even when dealing with a skewed variable, the Central Limit Theorem remains applicable as long as n is no less than 30.

For sums, this theorem can likewise be employed, accompanied by mean $\mu$ and standard deviation $s = \sqrt{n}*\sigma$ .

In this scenario, we are given that:

$\mu = 11.5, \sigma = 4$

a. For a group of 50 randomly selected customers, what is the estimated 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 derived from 1 minus the p-value of Z corresponding to X = 12.

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

According to the Central Limit Theorem

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

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

$Z = 0.88$

$Z = 0.88$ yields a p-value of 0.8106.

1 - 0.8106 = 0.1894

Therefore, there is an 18.94% chance that the sample mean amount purchased is at least 12 gallons.

b. For a group of 50 randomly selected customers, what is the estimated probability that the total amount of gasoline purchased does not exceed 600 gallons?

Regarding sums, we have $mu = 50*11.5 = 575, s = \sqrt{50}*4 = 28.28$

This probability equals the p-value of Z when X = 600. Hence,

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

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

$Z = 0.88$

$Z = 0.88$ displays a p-value of 0.8106.

Thus, there is an 81.06% chance that the total gasoline purchased will be 600 gallons or less.

c. What is the approximate figure for the 95th percentile regarding the total purchases by 50 randomly chosen customers?

This value corresponds to X when Z indicates a p-value of 0.95, which occurs at 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 95th percentile estimate for the total amount purchased by 50 randomly selected customers stands at 621.5 gallons.