650-318=332

332/2=166

## Related Questions

. In a study of air-bag effectiveness it was found that in 821 crashes of midsize cars equipped with air bags, 46 of the crashes resulted in hospitalization of the drivers (based on data from the highway Loss Data Institute). Using a 0.01 significance level, you need to test the claim that the air-bag hospitalization rate is lower than the 7.8% rate for crashes of mid-size cars equipped with automatic safety belts. What conclusion should you make

P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584

0.00885... < 0.01

The test statistic of 46 is significant

There is sufficient evidence to reject H₀ and accept H₁

Air bags are more effective as protection than safety belts

Step-by-step explanation:

821 crashes

46 hospitalisations where car has air bags

7.8% or 0.078 probability of hospitalisations in cars with automatic safety belts

α = 0.01 or 1% ← level of significance

One-tailed test

We are testing whether hospitalisations in cars with air bags are less likely than in a car with automatic safety belts;

The likelihood of hospitalisation in a car with automatic safety belts, we are told, is 7.8% or 0.078;

So we are testing if hospitalisations in cars with air bags is less than 0.078;

So, firstly:

Let X be the continuous random variable, the number of hospitalisations from a car crash with equipped air bags

X~B(821, 0.078)

Null hypothesis (H₀): p = 0.078

Alternative hypothesis (H₁): p < 0.078

According to the information, we reject H₀ if:

P(X ≤ 46 | X~B(821, 0.078)) < 0.01

To find P(X ≤ 46) or equally P(X < 47), it could be quite long-winded to do manually for this particular scenario;

If you are interested, the manual process involves using the formula for every value of x up to and including 46, i.e. x = 0, x = 1, x = 2, etc. until x = 46, the formula is:

You can find binomial distribution calculators online, where you input n (i.e. the number of trials or 821 in this case), probability (i.e. 0.078) and the test statistic (i.e. 46), it does it all for you, which gives:

P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584

Now, we need to consider if the condition for rejecting H₀ is met and recognise that:

0.00885... < 0.01

There is sufficient evidence to reject H₀ and accept H₁.

To explain what this means:

The test statistic of 46 is significant according to the 1% significance level, meaning the likelihood that only 46 hospitalisations are seen in car crashes with air bags in the car as compared to the expected number in car crashes with automatic safety belts is very unlikely, less than 1%, to be simply down to chance;

In other words, there is 99%+ probability that the lower number of hospitalisations in car crashes with air bags is due to some reason, such as air bags being more effective as a protective implement than the safety belts in car crashes.

Three friends — let’s call them X, Y , and Z — like to play pool (pocket billiards). There are some pool games that involve three players, but these people instead like to play 9-ball, which is a game between two players with the property that a tie cannot occur (there’s always a winner and a loser in any given round). Since it’s not possible for all three of these friends to play at the same time, they use a simple rule to decide who plays in the next round: loser sits down. For example, suppose that, in round 1, X and Y play; then if X wins, Y sits down and the next game is between X and Z. Question: in the long run, which two players square off against each other most often? Least often? So far what I’ve described is completely realistic, but now we need to make a (strong) simplifying assumption. In practice people get tired and/or discouraged, so the probability that (say) X beats Y in any single round is probably not constant in time, but let’s pretend it is, to get a kind of baseline analysis: let 0 < pXY < 1 be the probability that X beats Y in any given game, and define 0 < pXZ < 1 and 0 < pY Z < 1 correspondingly. Consider the stochastic process P that keeps track of

Step-by-step explanation:

(a) If the state space is taken as , the probability of transitioning from one state, say (XY) to another state, say (XZ) will be the same as the probability of Y losing out to X, because if X and Y were playing and Y loses to X, then X and Z will play in the next match. This probability is constant with time, as mentioned in the question. Hence, the probabilities of moving from one state to another are constant over time. Hence, the Markov chain is time-homogeneous.

(b) The state transition matrix will be:

where as stated in part (b) above, the rows of the matrix state the probability of transitioning from one of the states (in that order) at time n and the columns of the matrix state the probability of transitioning to one of the states (in the same order) at time n+1.

Consider the entries in the matrix. For example, if players X and Y are playing at time n (row 1), then X beats Y with probability , then since Y is the loser, he sits out and X plays with Z (column 2) at the next time step. Hence, P(1, 2) = . P(1, 1) = 0 because if X and Y are playing, one of them will be a loser and thus X and Y both together will not play at the next time step. , because if X and Y are playing, and Y beats X, the probability of which is, then Y and Z play each other at the next time step. Similarly,, because if X and Z are playing and X beats Z with probability, then X plays Y at the next time step.

(c) At equilibrium,

i.e., the steady state distribution v of the Markov Chain is such that after applying the transition probabilities (i.e., multiplying by the matrix P), we get back the same steady state distribution v. The Eigenvalues of the matrix P are found below:

The solutions are

These are the eigenvalues of P.

The sum of all the rows of the matrix is equal to 0 when Hence, one of the eigenvectors is :

The other eigenvectors can be found using Gaussian elimination:

Hence, we can write:

, where

and

After n time steps, the distribution of states is:

Let n be very large, say n = 1000 (steady state) and let v0 = [0.333 0.333 0.333] be the initial state. then,

Hence,

Now, it can be verified that

Find f(5). f(x) = x2 +2x

f(5) = 35

Step-by-step explanation:

In any function f(x), x is the input and f(x) is the output

Example:

f(x) = 2x + 3, find f(a)

That means we need to find value of f(x) at x = a, so substitute x by a

f(a) = 2(a) + 3 = 2a + 3

f(a) = 2a + 3

Let us use the same way to solve our question

f(x) = x² + 2x

→ We need to find f(5), so substitute x by 5

∴ f(5) = (5)² + 2(5)

∴ f(5) = 25 + 10

f(5) = 35

Assuming the weights are in kilograms, what is the mystery weight of y Use the equation you created in part A to find thesolution
Type the correct answer in the box. Use numerals instead of words. If necessary, use / for a fraction bar

y =  1/2  kilogram(s)

Step-by-step explanation:

Solve the equation created from the model for y:

...........................

b) b√a

Step-by-step explanation:

Given equation,

→ ab²

Then the square root of ab² is,

→ square root of ab²

→ √(ab²)

→ b√a

Hence, option (b) is the answer.

B) b√a

Step-by-step explanation:

b√a

The television show CSI: Shoboygan has been successful for many years. That show recently had a share of 18, meaning that among the TV sets in use, 18% were tuned to CSI: Shoboygan. Assume that an advertiser wants to verify that 18% share value by conducting its own survey, and a pilot survey begins with 14 households have TV sets in use at the time of a CSI: Shoboygan broadcast. Find the probability that none of the households are tuned to CSI: Shoboygan. P(none) = Find the probability that at least one household is tuned to CSI: Shoboygan. P(at least one) = Find the probability that at most one household is tuned to CSI: Shoboygan. P(at most one) = If at most one household is tuned to CSI: Shoboygan, does it appear that the 18% share value is wrong? (Hint: Is the occurrence of at most one household tuned to CSI: Shoboygan unusual?)

(a) The value of P (None) is 0.062.

(b) The value of P(at least one) is 0.938.

(c) The value of P(at most one) is 0.253.

(d) The event is not unusual.

Step-by-step explanation:

Let X = number of households watching the show.

The probability of the random variable x is, P (X) = p = 0.18.

The sample selected for the survey is of size, n = 14

The random variable X follows a Binomial distribution with parameter n = 14 and p = 0.18.

The probability of a Binomial distribution is computed using the formula:

(a)

Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:

Thus, the value of P (None) is 0.062.

(b)

Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

Thus, the value of P(at least one) is 0.938.

(c)

Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

Thus, the value of P(at most one) is 0.253.

(d)

An event that has a very low probability of occurrence is known as an unusual event.

The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.

This probability value is not low.

Hence, the event is not unusual.