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charlie builds decks for a living. he builds 14 standard decks in 3 weeks. how many decks can he build in 3 months (12 weeks)?

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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

babunello [11817]

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.

5 0

2 months ago

An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th

tester [12383]

Answer:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$

Step-by-step explanation:

Notation

$\bar X$ is the sample mean

$\mu$ indicates the population mean (the variable of interest)

s signifies the sample standard deviation

n denotes the sample size

Solution to the problem

The mean's confidence interval is derived from the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In this instance, the 9% confidence interval corresponds to:

$8.8104 \leq \mu \leq 11.1248$

We can determine the mean using the following:

$\bar X = \frac{8.8104 +11.1248}{2}= 9.9676$

Additionally, the margin of error can be calculated as:

$ME= \frac{11.1248- 8.8104}{2}= 1.1572$

The margin of error for this situation is expressed as:

$ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE$

Next, we find the standard error:

$SE = \frac{ME}{t_{\alpha/2}}$

The critical value for a 95% confidence interval using a normal standard distribution is roughly 1.96, and substituting gives us:

$SE = \frac{1.1572}{1.96}= 0.5904$

For the 98% confidence interval, the significance corresponds to $\alpha=1-0.98= 0.02$ and $\alpha/2 = 0.01$ with a critical value of 2.326, yielding a confidence interval of:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$

8 0

2 months ago

The hourly wages earned by 20 employees are shown in the first box-and-whisker plot below. The person earning $15 per hour quits

Leona [12618]

Answer:

The mean decreases, while the median stays unchanged.

Step-by-step explanation:

A box plot is constructed from the quartiles of the data distribution, incorporating the maximum and minimum values. From this, one can determine the range, median, and interquartile range.

In this scenario, the median maintains its value at $9.5 per hour. The median is represented by the middle line in the box, which remains constant.

Conversely, the range of this data set shrinks from 7 to 3.

Moreover, the mean must decrease because values greater than $11 are no longer represented in the second box plot. The mean is a measure sensitive to such alterations.

Conclusively, the correct finding is The mean decreases, while the median remains unchanged.

8 0

3 months ago

Engineers measure angles in gradients, which are smaller than degrees. The table shows the conversion of some angle measures in

tester [12383]

The following table presents the conversion from degrees to gradients.

To calculate the slope, we take the difference between the two y-values (gradients) and divide it by the difference between the corresponding x-values (degrees).

For this purpose, we will use the initial and final points listed in the table. Therefore, the slope m is calculated as:

$m= \frac{300-(-200)}{270-(-180)} \\ \\ 
m= \frac{500}{450} \\ \\ 
m=1.11$

After rounding to two decimal places, the slope of the line converting degrees to gradients is 1.11

8 0

3 months ago

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