Plzzz helpppp itssss urgentt!!!!!!!!

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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".

Part a

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

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

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:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

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:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

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:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

5 0

3 years ago

Help! i’m so confused.

3241004551 [841]

Ok what happened what do you need help on

8 0

4 years ago

Read 2 more answers

Tomas has $1,000 to spend on a vacation. His

Jobisdone [24]

Answer:

He can spend $118.50 each day

Step-by-step explanation:

1000 - 348.25 = 651.75

651.75 / 5.5 = 118.5

4 0

3 years ago

Dinesh has been keeping track of his time at work over the past month. Answer the questions below using the information provided

Gala2k [10]

The total hours that Dinesh worked each week over the past month are as follows:

<h3>Dinesh's Time in Hours</h3>

Total Hours

Worked

Week 1 37.500

Week 2 36.071

Week 3 35.999

Week 4 34.570

Total 144.140

<h3>Data and Calculations:</h3><h3>Dinesh's Time in Hours</h3>

Meetings Administration Project X - Project X- Total Hours

Design Deliverables Worked

Week 1 7¹/₄ 4¹/₄ 23¹/₃ 2²/₃

Week 2 4²/₃ 2⁴/₇ 16¹/₃ 12¹/₂

Week 3 3.50 3.83 9.33 19.33

Week 4 6.17 3.40 4.20 20.80

<h3>Conversion into Dinesh's Time in Hours (decimals)</h3>

Meetings Administration Project X - Project X- Total Hours

Design Deliverables Worked

Week 1 7.250 4.250 23.333 2.667 37.500

Week 2 4.667 2.571 16.333 12.500 36.071

Week 3 3.500 3.833 9.333 19.333 35.999

Week 4 6.170 3.400 4.200 20.800 34.570

Total 21.587 14.054 53.199 55.300 144.140

Thus, Dinesh's total hours worked for the past month was <u>144.140 hours</u>.

Learn more about calculating hours worked at brainly.com/question/17054097

#SPJ1

8 0

2 years ago

John is playing a game with a standard deck of playing cards. He wants to draw a jack on the first try. Which of the following s

Tatiana [17]

Answer:

d

Step-by-step explanation:

It is a 1/13 chance but even if you shuffle and replace it, its still a 1/13 chance.

4 0

3 years ago

Read 2 more answers