Concerns about climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g., from renewable sources) and petrodiesel (from fossil fuel). Random samples of 32 blended fuels are tested in a lab to ascertain the bio/total carbon ratio. (a) If the true mean is .9370 with a standard deviation of 0.0090, within what interval will 99 percent of the sample means fall

Answer:

99% of the sample means will fall between 0.93288 and 0.94112.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The true mean is .9370 with a standard deviation of 0.0090

This means that [tex]\mu = 0.9370, \sigma = 0.0090[/tex]

Sample of 32:

This means that [tex]n = 32, s = \frac{0.009}{32} = 0.0016[/tex]

Within what interval will 99 percent of the sample means fall?

Between the 50 - (99/2) = 0.5th percentile and the 50 + (99/2) = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]-2.575 = \frac{X - 0.9370}{0.0016}[/tex]

[tex]X - 0.9370 = -2.575*0.0016[/tex]

[tex]X = 0.93288[/tex]

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]2.575 = \frac{X - 0.9370}{0.0016}[/tex]

[tex]X - 0.9370 = 2.575*0.0016[/tex]

[tex]X = 0.94112[/tex]

99% of the sample means will fall between 0.93288 and 0.94112.

Use the substitution method for solve for x and y.


y = 2x + 10
y = -4x - 2

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