Can u plz help me with this

Of all the registered automobiles in a city, 12% fail the emissions test. Fourteen automobiles are selected at random to undergo

postnew [5]

Answer:

a) 0.1542

b) 0.7685

c) 0.2315

d) No, it is not unusual

Explanation:

The procedure to make the test meets the requirements of binomial experiments because:

there are two possible mutually exclusive outputs: fail the test, or pass the test.
the probability of each event remains constant during all the test (p=12% = 0.12, for failing the test, and 1-p = 88% = 0.88, for passing the test)
each trial (test) is independent of other trial.

Solution

(a) Find the probability that exactly three of them fail the test.

You want P(X=3)

Using the equation for discrete binomial experiments, the probability of exactly x successes is:

$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{(n-x)}$

Substituting C(n,x) with its developed form, that is:

$P(X=x)=\dfrac{n!}{x!\cdot (n-x)!}\cdot p^x\cdot (1-p)^{(n-x)}$

Thus, you must use:

x = 3 (number of automobiles that fail the emissions test)
n = 14 (the number of automobiles selected to undergo the emissions test),
p = 0.12 (probability of failing the test; this is the success of the variable on our binomial experiment)
1 - p = 0.88 (probability of passing the test; this is the fail of the variable on our binomial experiment)

$P(X=3)=\dfrac{14!}{3!\cdot (14-3)!}\cdot 0.12^3\cdot 0.88^{11}=0.1542$

(b) Find the probability that fewer than three of them fail the test.

The probability that fewer than three of them fail the test is the probability that exactly 0, or exactly 1, or exactly 2 fail the test.

That is: P(X=0) + P(X=1) + P(X=2)

Using the same formula:

$P(X=0)=\dfrac{14!}{0!\cdot 14!}\times 0.12^0\cdot 0.88^{14}$

$P(X=0)=0.1670$

$P(X=1)=\dfrac{14!}{1!\cdot 13!}\cdot 0.12^1\cdot 0.88^{13}$

$P(X=1)=0.3188$

$P(X=2)=\dfrac{14!}{2!\cdot 12!}\codt0.12^2\cdot 0.88^{12}$

$P(X=2)=0.2826$

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.7685

(c) Find the probability that more than two of them fail the test.

The probability that more than two of them fail the test is equal to 1 less the probability that exactly 0, or exactly 1, or exactly 2 fail the test:

P( X > 2) = 1 - P( X = 0) - P(X = 1) - P(X = 2)

P X > 2) = 1 - [P(X=0) + P(X = 1) + P(X = 2)]

P (X > 2) = 1 - [0.7685]

P (X > 2) = 0.2315

(d) Would it be unusual for none of them to fail the test?

Remember that not failing the test is the fail of the binomial distribution. Thus, none of them failing the test is the same as all of them passing the test.

You can find the probability that all the automibles pass the emission tests by multiplying the probability of passing the test (0.88) 14 times.

Then, the probability that none of them to fail the test is equal to:

$(1-p)^{14}\\\\(0.88)^{14}=0.1671$

That means that the probability than none of the automobiles of the sample fail the test is 16.71%.

Unusual events are usually taken as events with a probability less than 5%. Thus, this event should not be considered as unusual.

5 0

3 years ago

Please help asap i need to pass this!

mr Goodwill [35]

Answer:

B

Step-by-step explanation:

i had that same question to answer and got it right

6 0

3 years ago

I have a Triangle and I need to find the side BC?

Rus_ich [418]

Looking at this problem in the book, I'm guessing that you've been
introduced to a little bit of trigonometry. Or at least you've seen the
definitions of the trig functions of angles.

Do you remember the definition of either the sine or the cosine of an angle ?

In a right triangle, the sine of an acute angle is (opposite side) / (hypotenuse),
and the cosine of an acute angle is (adjacent side) / (hypotenuse).

Maybe you could use one of these to solve this problem, but first you'd need to
make sure that this is a right triangle.

Let's see . . . all three angles in any triangle always add up to 180 degrees.
We know two of the angles in this triangle ... 39 and 51 degrees.
How many degrees are left over for the third angle ?
180 - (39 + 51) = 180 - (90) = 90 degrees for the third angle.
It's a right triangle ! yay ! We can use sine or cosine if we want to.

Let's use the 51° angle.
The cosine of any angle is (adjacent side) / (hypotenuse) .
'BC' is the side adjacent to the 51° angle in the picture,
and the hypotenuse is 27 .

cosine(51°) = (side BC) / 27

Multiply each side of that equation by 27 :

Side-BC = (27) times cosine(51°)

Look up the cosine of 51° in a book or on your calculator.

Cosine(51°) = 0.62932 (rounded)

Side BC = (27) x (0.62932) = 16.992 (rounded)
============================================

You could just as easily have used the sine of 39° .
That would be (opposite side) / (hypotenuse) ... also (side-BC) / 27 .

4 0

4 years ago

2 MATH K.

Ann [662]

Answer:

94

Step-by-step explanation:

Given the addition problem:

58 + 36

We can form a table :

5 _______ 8

+

3 _______ 6

9 4

Hence, (58 + 36) = 94

4 0

3 years ago

What are the independent and dependent quantities?

Anit [1.1K]

There are two types of variables-independent and dependent. Answer: An independent variable is exactly what it sounds like. It is a variable that stands alone and isn't changed by the other variables you are trying to measure. For example, someone's age might be an independent variable.

Hope this helps! :)

5 0

3 years ago

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