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The probability distribution for the random variable, \(x\), is \begin{tabular}{lllllll} \hline\(x\) & 2 & \(2.5\) & \(3.0\) & \(3.5\) & \(4.0\) & \(4.5\) \\ \(P(x)\) & \(0.07\) & \(0.36\) & \(0.21\) & \(0.19\) & \(0.10\) & \(0.07\) \\ \hline \end{tabular} (a) State \(P(x=3.5)\) (b) Calculate \(P(x \geqslant 3.0)\) (c) Calculate \(P(x<4.0)\) (d) Calculate \(P(x>3.5)\) (e) Calculate \(P(x \leqslant 3.9)\) (f) The variable, \(x\), is sampled 50000 times. How many times would you expect \(x\) to have a value of \(2.5 ?\)

Short Answer

Expert verified
a) 0.19, b) 0.57, c) 0.93, d) 0.17, e) 0.93, f) 18000

Step by step solution

01

Identify P(x=3.5)

Look at the probability distribution table and find the probability associated with the random variable x when x equals 3.5.
02

Calculate P(x \geqslant 3.0)

Add up the probabilities in the table for the values of x that are greater than or equal to 3.0.
03

Calculate P(x

Add up the probabilities in the table for the values of x that are less than 4.0.
04

Calculate P(x>3.5)

Add up the probabilities in the table for the values of x that are strictly greater than 3.5.
05

Calculate P(x \leqslant 3.9)

Since all of the values of x are less than or equal to 3.9 (under 4.0), this is the same as P(x<4.0). Use the result from Step 3.
06

Expectation of x=2.5

Multiply the probability of x being 2.5 by the number of times x is sampled (50000) to find the expected number of times x will equal 2.5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variable
In the realm of probability, a random variable is a fundamental concept that represents a quantity whose value is subject to varying outcomes of a random phenomenon. Imagine tossing a coin; the result can be either head or tail. The random variable in this scenario could represent the number of heads you get in a series of tosses. The values a random variable can take are associated with different probabilities.

In the exercise, the variable 'x' is a discrete random variable since it has specific separate values it can take, such as 2, 2.5, 3.0, and so on. The table is essentially a probability distribution that tells us how probable each outcome is. Understanding this helps you accurately determine the likelihood of any given occurrence of 'x'.
Expected Value Calculation
The expected value is a measure that gives us an idea of the 'average' outcome we can anticipate from a random variable over many trials. It takes into account all the possible values of the random variable and their respective probabilities. To calculate it, we multiply each possible value of the variable by its probability and sum up these products.

For example, if a dice is rolled, the expected value of the outcome would be calculated as: \( (1 \times \frac{1}{6}) + (2 \times \frac{1}{6}) + (3 \times \frac{1}{6}) + (4 \times \frac{1}{6}) + (5 \times \frac{1}{6}) + (6 \times \frac{1}{6}) \), which equals 3.5.

In our exercise, to find the expected number of times 'x' will equal 2.5 when sampled 50000 times, we use the provided probability of 0.36 and multiply it by 50000, which gives us an expected count of 18000 times.
Probability Calculations
Probability calculations are used to quantify the likelihood of events represented by a random variable using a scale from 0 to 1, where 0 denotes impossibility and 1 denotes certainty. To solve various probability questions for a discrete random variable, you sum the probabilities of the outcomes that satisfy the condition in question.

For instance, for part (a) 'P(x=3.5)', one simply looks up the value directly in the distribution table. In more compound situations, like calculating 'P(x \geqslant 3.0)', you need to add the probabilities for all 'x' values that are equal to or greater than 3.0.

It's important to read the inequalities correctly: 'greater than or equal to' (\( \geqslant \) ) and 'less than' (\( < \)) indicate which values to include in your calculation. The sums of these probabilities provide answers to the posed queries regarding the behavior of the random variable 'x'.

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Most popular questions from this chapter

A machine manufactures electrical components for the car industry at the rate of 750 per hour. The probability a component is faulty is \(0.013\). Use both the binomial distribution and the corresponding Poisson approximation to find the probability that in a sample of 200 components (a) none are faulty (b) one is faulty (c) two are faulty (d) three are faulty (e) more than three are faulty

The probability of a disk drive failure in any week is 0.007. A computer service company maintains 900 disk drives. Use the Poisson distribution to calculate the probability of (a) seven (b) more than seven disk drive failures in a week.

\(f(x)=k x^{2}, k\) constant, \(-1 \leqslant x \leqslant 1 . f(x)\) is a p.d.f. (a) What is the value of \(k\) ? (b) Calculate the probability that \(x>0.5\). (c) If \(P(x>c)=0.6\) then what is the value of \(c ?\)

A random variable, \(x\), has a standard normal distribution. Calculate the probability that \(x\) lies in the following intervals: (a) \((0.25,0.75)\) (b) \((-0.3,0.1)\) (c) within \(1.5\) standard deviations of the mean (d) more than two standard deviations from the mean (e) \((-1.7,-0.2)\)

A machine requires all seven of its micro-chips to operate correctly in order to be acceptable. The probability a micro-chip is operating correctly is \(0.99\). (a) What is the probability the machine is acceptable? (b) What is the probability that six of the seven chips are operating correctly? (c) The machine is redesigned so that the original seven chips are replaced by four new chips. The probability a new chip operates correctly is \(0.98\). Is the new design more or less reliable than the original?

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