Im stuck lol, I'm doing algebra 1, the answer is supposed to be
%20" id="TexFormula1" title="27 \frac{3}{8} " alt="27 \frac{3}{8} " align="absmiddle" class="latex-formula">
1 answer:
You might be interested in
Answer:
A truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.
For a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.
Step-by-step explanation:
We have here a <em>random variable</em> <em>normally distributed</em> (the time between engine failures). According to this, most values are around the mean of the distribution and less are far from it considering both extremes of the distribution.
The <em>normal distribution</em> is defined by two parameters: the population mean and the population standard deviation, and we have each of them:
miles.
miles.
To find the probabilities asked in the question, we need to follow the next concepts and steps:
- We will use the concept of the <em>standard normal distribution</em>, which has a mean = 0, and a standard deviation = 1. Why? With this distribution, we can easily find the probabilities of any normally distributed data, after obtaining the corresponding <em>z-score</em>.
- A z-score is a kind of <em>standardized value</em> which tells us the <em>distance of a raw score from the mean in standard deviation units</em>. The formula for it is:
. Where <em>x</em> is the value for the raw score (in this case x = 5000 miles).
- The values for probabilities for the standard normal distribution are tabulated in the <em>standard normal table</em> (available in Statistics books and on the Internet). We will use the <em>cumulative standard normal table</em> (see below).
With this information, we can solve the first part of the question.
The chance that a truck will be able to travel a total distance of over 5000 miles without an engine failure
We can "translate" the former mathematically as:
miles.
The z-score for x = 5000 miles is:
This value of z is negative, and it tells us that the raw score is 1.25 standard deviations <em>below</em> the population mean. Most standard normal tables are made using positive values for z. However, since the normal distribution is symmetrical, we can use the following formula to overcome this:
So
Consulting a standard normal table available on the Internet, we have
Then
However, this value is for P(z<-1.25), and we need to find the probability P(z>-1.25) = P(x>5000) (Remember that we standardized x to z, but the probabilities are the same).
In this way, we have
That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:
In words, a truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.
We can see the former probability in the graph below.
The chance that a fleet of a dozen trucks will have an average time-between-failures of 5000 miles or more
We are asked here for a sample of <em>12 trucks</em>, and this is a problem of <em>the sampling distribution of the means</em>.
In this case, we have samples from a <em>normally distributed data</em>, then, the sample means are also normally distributed. Mathematically:
In words, the samples means are normally distributed with the same mean of the population mean , but with a standard deviation
.
We have also a standardized variable that follows a standard normal distribution (mean = 0, standard deviation = 1), and we use it to find the probability in question. That is
Then
The "average time-between-failures of 5000" is . In other words, this is the mean of the sample of the 12 trucks.
Thus
This value is so low for z, that it tells us that P(z>-4.33) is almost 1, in other words it is almost certain that for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is almost 1.
The complement of P(z<-4.33) is:
or practically 1.
In conclusion, for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.
