Can i have some help

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

Find the first five terms of the sequence in which a1 = –10 and an = 4an – 1 + 7, if n ≥ 2.

Naddika [18.5K]

Answer:

-10, -33, -125, -493, -1965

Step-by-step explanation:

a1 is -10, meaning the first term is -10. therefore the first term of the sequence has to be -10

7 0

3 years ago

How many deaths per day if approximately 32200 citizens of a country died in automobile accidents in 2015.

Stolb23 [73]

32200 divide by the number of days in a year 365

32200 / 365 = 88.21

about 88 per day

4 0

2 years ago

Read 2 more answers

15% a football originally priced at $18.00

lesantik [10]

Answer:

2.7

Step-by-step explanation:

15% of 18.00 is 2.7

7 0

3 years ago

Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twenty-five years later (in 2008) that perc

Dima020 [189]

Answer:

a) $ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

b) $ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.96$

If we replace the values into equation (a) for 1983 we got:

$ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019$

Part b

Since is the same confidence level the z value it's the same.

If we replace the values into equation (a) for 2008 we got:

$ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245$

Is the margin of error the same in parts (a) and (b)? Why or why not?

On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of $\hat p$ the margin of error change for part a and b.

3 0

3 years ago