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3 years ago

Below are 36 sorted ages of an acting award winner. Find Upper P70 using the method presented in the textbook.

Mathematics

2 answers:

Ipatiy [6.2K]3 years ago

8 0

58 is your answer. hope it helps

Leona [35]3 years ago

6 0

The upper 70th percentile is the number below which 70% of the data lie.

The 70th percentile position is given by:

$P_{70}= \frac{70N}{100} = \frac{70(36)}{100} =25.2$

Thus, the 70th percentile position is approximately the 25th data item (after the data has been arranged in acsending order).

Given the following data:
<span>16 24 25 26 27 29 36 39 39 39 40 44 45 47 47 48 50 51 51 53 53 54 57 58
58 60 65 66 67 69 69 71 72 74 74 74

The 25th data in the data set is 58.

Therefore, the upper P70 is 58.
</span>

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A manager at a company that manufactures cell phones has noticed that the number of faulty cell phones in a production run of ce

ExtremeBDS [4]

Answer:

a) Poisson probability distribution

b) The probability of no faulty cell phones will be produced tomorrow is 0.1653

c) The probability of 3 or more faulty cell phones were produced in today's run is 0.2694

Step-by-step explanation:

Poisson distribution is used for independent events which occur at a constant rate within a given interval of time

The Poisson probability distribution formula is P(x; μ) = (e^-μ) (μ^x) / x! where

x is the actual number of successes that result from the experiment

e is approximately equal to 2.71828

μ is the mean of the distribution

The number of faulty cell phones in a production run of cell phones is usually small and that the quality of one day's run seems to have no bearing on the next day

That means the probability of finding faulty phones in first day not depend on finding another days

Then the model might you use to model the number of faulty cell phones produced in one day is Poisson probability distribution

a) Poisson probability distribution

∵ The mean number of faulty cell phones is 1.8 per day

∴ μ = 1.8

∵ There is no faulty cell phones will be produced tomorrow

∴ x = 0

- Use the formula above to find the probability

∵ P(0 ; 1.8) = (e^-1.8) (1.8^0) / 0!

- Remember 1.8^0 = 1 and 0! = 1

∴ P(0 ; 1.8) = (e^-1.8)(1)/(1)

∴ P(0 ; 1.8) = 0.1653

b) The probability of no faulty cell phones will be produced tomorrow is 0.1653

∵ The mean number of faulty cell phones is 1.8 per day

∴ μ = 1.8

∵ There is 3 or more faulty cell phones were produced in today's run

∴ x ≥ 3

∵ P(x ≥ 3) = 1 - P(x = 0) - P(x = 1) - P(x = 2)

- Let us find P(1 ; 1.8) and P(2 ; 1.8)

∵ P(1 ; 1.8) = (e^-1.8) (1.8^1) / 1!

∵ 1.8^1 = 1.8

∵ 1! = 1

∴ P(1 ; 1.8) = (e^-1.8)(1.8)/(1)

∴ P(1 ; 1.8) = 0.2975

∵ P(2 ; 1.8) = (e^-1.8) (1.8^2) / 2!

∵ 1.8^2 = 3.24

∵ 2! = 2

∴ P(2 ; 1.8) = (e^-1.8)(3.24)/(2)

∴ P(2 ; 1.8) = 0.2678

Substitute them in the rule above

∵ P(x ≥ 3 ; 1.8) = 1 - 0.1653 - 0.2975 - 0.2678

∴ P(x ≥ 3 ; 1.8) = 0.2694

c) The probability of 3 or more faulty cell phones were produced in today's run is 0.2694

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Strike441 [17]

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3 years ago

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Sphinxa [80]

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3 years ago

A man travels 20 km by car from Town P to Town Q at an average speed of x km/h. He finds that the time of the journey would be s

yuradex [85]

Answer:

x = 20.

Step-by-step explanation:

First, you should remember the relation:

Distance = Speed*Time.

First, we know that a man travels a distance of 20km at a speed of x km/h, in a time T.

We can write this as:

20km = (x km/h)*T

We know that the time is shortened by 12 minutes if the speed is increased by 5km/h

Rewriting these 12 minutes in hours (remember that 60min = 1 hour)

12 min = (12/60) hours = 0.2 hours

Then from this, he can travel the same distance of 20km in a time T minus 0.2 hours if the speed is increased by 5 km/h

We can write this as:

20km = (x + 5 km/h)*(T - 0.2 h)

Then we have a system of two equations, and we want to find the value of x:

20km = (x km/h)*T

20km = (x + 5 km/h)*(T - 0.2 h)

First, we should isolate the variable T in one of the equations, if we isolate it in the first one, we will get:

20km/(x km/h) = T

Replacing that in the other equation we get:

20km = (x + 5 km/h)*(T - 0.2 h)

20km = (x + 5 km/h)*( 20km/(x km/h) - 0.2 h)

Now we can solve this for x.

Removing the units (that we know that are correct) so the math is easier to read, we get:

20 = (x + 5)*(20/x - 0.2)

We only want to solve this for x.

20 = x*20/x - x*0.2 + 5*20/x - 5*0.2

20 = 20 - 0.2*x + 100/x - 1

subtracting 20 in both sides we get:

20 - 20 = 20 - 0.2*x + 100/x - 1 - 20

0 = -0.2*x + 100/x - 1

If we multiply both sides by x we get:

0 = -0.2*x^2 + 100 - x

-0.2*x^2 - x + 100 = 0

This is just a quadratic equation, we can solve it using the Bhaskara's equation, the solutions are:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*(-0.2)*100} }{2*-0.2} = \frac{1 \pm 9 }{-0.4}$

Then the two solutions are:

x = (1 + 9)/-0.4 = -25

x = (1 - 9)/-0.4 = 20

As x is used to represent a speed, the negative solution does not make sense, so we should use the positive one.

x = 20

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2 years ago