1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Reil [10]
3 years ago
6

2. The time between engine failures for a 2-1/2-ton truck used by the military is

Mathematics
1 answer:
OLEGan [10]3 years ago
7 0

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:

\\ \mu = 6000 miles.

\\ \sigma = 800 miles.

To find the probabilities asked in the question, we need to follow the next concepts and steps:

  1. 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>.
  2. 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: \\ z = \frac{x - \mu}{\sigma}. Where <em>x</em> is the value for the raw score (in this case x = 5000 miles).
  3. 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:

\\ P(x>5000) miles.

The z-score for x = 5000 miles is:

\\ z = \frac{5000 - 6000}{800}

\\ z = \frac{-1000}{800}

\\ z = -1.25

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:

\\ P(z

So

\\ P(z

Consulting a standard normal table available on the Internet, we have

\\ P(z

Then

\\ P(z1.25)

\\ P(z1.25)

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

\\ P(z>-1.25) = 1 - P(z

That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:

\\ P(z>-1.25) = 1 - 0.10565

\\ P(z>-1.25) = 0.89435  

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:

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

In words, the samples means are normally distributed with the same mean of the population mean \\ \mu, but with a standard deviation \\ \frac{\sigma}{\sqrt{n}}.

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

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z \sim N(0, 1)

Then

The "average time-between-failures of 5000" is \\ \overline{x} = 5000. In other words, this is the mean of the sample of the 12 trucks.

Thus

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z = \frac{5000 - 6000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{230.940148}

\\ z = -4.330126

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.

\\ P(z

\\ P(z

\\ P(z

The complement of P(z<-4.33) is:

\\ P(z>-4.33) = 1 - P(z 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.

You might be interested in
How do I find the midpoint
Hunter-Best [27]

To find the midpoint of a line use the midpoint formula which is M = (x1+x2/2 , y1 +y2/2) After plugging in the coordinates for your variables you should get (-1+5/2, -1+2/2) This simplifies to (2, 1/2) which is the coordinates for the middle point of the base XY.

5 0
3 years ago
(02.02 MC) Triangle ABC is shown. A is at negative 2, 1. B is at negative 1, 4. C is at negative 4, 5. If triangle ABC is reflec
Bezzdna [24]

Given:

The vertices of a triangle ABC are A(-2,1), B(-1,4) and C(-4,5).

Triangle ABC is reflected over the x‐axis, reflected over the y‐axis, and rotated 180 degrees.

To find:

The point B'.

Solution:

If a figure reflected over x-axis, then

(x,y)\to (x,-y)

B(-1,4)\to B_1(-1,-4)

If a figure reflected over y-axis, then

(x,y)\to (-x,y)

B_1(-1,-4)\to B_2(-(-1),-4)

B_1(-1,-4)\to B_2(1,-4)

If a figure rotated 180 degrees about the origin, then

(x,y)\to (-x,-y)

B_2(1,-4)\to B'(-(1),-(-4))

B_2(1,-4)\to B'(-1,4)

So, the coordinate of point B' are (-1,4).

Therefore, the correct option is A.

6 0
3 years ago
PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ HELPPPPPPPPPPPPPPPPPPP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
MaRussiya [10]

speed =  \frac{distance}{time}  \\

7 0
2 years ago
Renna pushes the elevator button, but the elevator does not move. The mass limit for the elevator is 450 kilograms (kg), but Ren
liraira [26]

The picture below has both of the answers to your problem.

7 0
3 years ago
Read 2 more answers
A six-sided die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on th
agasfer [191]

Answer:

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

Step-by-step explanation:

Given that; the probability of each face turning up is proportional to the number of dots on that face

P(1) = 1×P(1)

P(2) = 2×P(1)

P(3) = 3×P(1)

P(4) = 4×P(1)

P(5) = 5×P(1)

P(6) = 6×P(1)

P(T) = 21×P(1)

Where;

P(x) is the probability of getting number x on the dice.

P(T) is the total probability of obtaining any number

N(x) is the number of possible number x in terms of the distribution function.

P(x) = N(x)/N(T) ....1

And since P(T) is constant, and P(T) is proportional to N(T) then,

P(x) is directly proportional to N(x)

So, equation 1 becomes;

P(x) = N(x)/N(T) = P(x)/P(T) ....2

The probability of getting either a 5 or a 2 in one throw

P(2U5) = (P(2) + P(5))/P(T)

Substituting the values of each probability;

P(2U5) = (2P(1) + 5P(1))/21P(1)

P(2U5) = 7P(1)/21P(1)

P(1) cancel out, to give;

P(2U5) = 7/21 = 1/3

the probability of getting either a 5 or a 2 in one throw is 1/3

8 0
3 years ago
Other questions:
  • The number w and 0.8 are additive inverses.
    11·2 answers
  • 4. En el Vivero de Arturo hay 226
    7·1 answer
  • If nondisjunction occurs in meiosis ii during gametogenesis, what will be the result at the completion of meiosis? a all the gam
    10·1 answer
  • I need help, giving brainliest.
    6·2 answers
  • What is the 8th term of a(n)=6•3^(n-1)
    5·1 answer
  • Please Help me with number two and three
    8·2 answers
  • Describe the graph of function g by observing the graph of the base function f.
    7·2 answers
  • 13. I have 22 nickels, dimes and quarters worth $3.45. There are twice as many quarters as nickels. Find the number of each coin
    5·1 answer
  • Pls explain ASAP <br> Thanks
    9·2 answers
  • Choose all numbers that are divisible by 3
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!