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3 years ago

15

If a trucker changes the oil every 10,000 miles, and he starts the year with a fresh oil change, how many times will he change t

he oil in a year? PLZ HELP ME ITS DUE TOMORROW AND PLZZ SHOW YOUR WORK

2 answers:

Wittaler [7]3 years ago

8 0

he will change his oil 16 times because 160,000 divided by 10,000 equals 16

Rasek [7]3 years ago

5 0

Answer:

2times

Step-by-step explanation:

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3 years ago

Membership in Mensa requires an IQ score above 131.5. Nine candidates take an IQ test, and they have a mean IQ score of 133. IQ

rodikova [14]

Answer:

a) $P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

$P(Z\geq 2.2)=1- P(z$

b) $P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

And we can find this probability using the complement rule:

$P(Z\geq 6.6)=1- P(z$

c) No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

$P(Z\leq 6.3) \approx 1$

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

A. If 1 person is randomly selected from thegeneral population, find the probability of getting someone with anIQ score of at least 133.

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(100,15)$

Where $\mu=100$ and $\sigma=15$

We are interested on this probability

$P(X\geq 133)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

$P(Z\geq 2.2)=1- P(z$

And the probability is calculated from the normal standard table or with excel.

B. If 9 people are randomly selected,find the probability that their mean IQ score is at least 133.

For this case since the distribution for the random variable X is normal then the distribution for the sample mean is also normal and given by:

$\bar X = \sim N(\mu= 100 ,\sigma_{\bar x}= \frac{15}{\sqrt{9}}=5)$

The new z score is defined as

$z=\frac{x-\mu}{\sigma_{\bar x}}$

If we apply this formula to our probability we got this:

$P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

And we can find this probability using the complement rule:

$P(Z\geq 6.6)=1- P(z$

And the probability is calculated from the normal standard table or with excel.

C. Although the results are available,the individual IQ test scores have been lost. Can it be concluded that all 9 candidates have IQ scores above 131.5 so that they allare eligible for Mensa membership?

No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

$P(Z\leq 6.3) \approx 1$

6 0

3 years ago

Read 2 more answers