Variance Of A Discrete Random Variable
Example 4A service organization in a large town organizes a raffle each month. One thousand raffle tickets are sold for $1 each. Each has an equal chance of winning. First prize is $300, second prize is $200, and third prize is $100. Let X denote the net gain from the purchase of one ticket.
Construct the probability distribution of X. Find the probability of winning any money in the purchase of one ticket. Find the expected value of X, and interpret its meaning.Solution:.If a ticket is selected as the first prize winner, the net gain to the purchaser is the $300 prize less the $1 that was paid for the ticket, hence X = 300 − 1 = 299. There is one such ticket, so P(299) = 0.001. Applying the same “income minus outgo” principle to the second and third prize winners and to the 997 losing tickets yields the probability distribution: x 299 199 99 − 1 P ( x ) 0.001 0.001 0.001 0.997.Let W denote the event that a ticket is selected to win one of the prizes. Using the table P ( W ) = P ( 299 ) + P ( 199 ) + P ( 99 ) = 0.001 + 0.001 + 0.001 = 0.003.Using the formula in the definition of expected value, E ( X ) = 299 0.001 + 199 0.001 + 99 0.001 + ( − 1 ) 0.997 = − 0.4The negative value means that one loses money on the average.
In particular, if someone were to buy tickets repeatedly, then although he would win now and then, on average he would lose 40 cents per ticket purchased.The concept of expected value is also basic to the insurance industry, as the following simplified example illustrates. Example 5A life insurance company will sell a $200,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $195. Find the expected value to the company of a single policy if a person in this risk group has a 99.97% chance of surviving one year.Solution:Let X denote the net gain to the company from the sale of one such policy. There are two possibilities: the insured person lives the whole year or the insured person dies before the year is up. Applying the “income minus outgo” principle, in the former case the value of X is 195 − 0; in the latter case it is 195 − 200,000 = − 199,805. Since the probability in the first case is 0.9997 and in the second case is 1 − 0.9997 = 0.0003, the probability distribution for X is: x 195 − 199,805 P ( x ) 0.9997 0.0003Therefore E ( X ) = Σ x P ( x ) = 195 0.9997 + ( − 199,805 ) 0.0003 = 135Occasionally (in fact, 3 times in 10,000) the company loses a large amount of money on a policy, but typically it gains $195, which by our computation of E ( X ) works out to a net gain of $135 per policy sold, on average.
DefinitionThe The number Σ ( x − μ ) 2 P ( x ) (also computed using Σ x 2 P ( x ) − μ 2), measuring its variability under repeated trials., σ, of a discrete random variable X is the square root of its variance, hence is given by the formulas σ = Σ ( x − μ ) 2 P ( x ) = Σ x 2 P ( x ) − μ 2The variance and standard deviation of a discrete random variable X may be interpreted as measures of the variability of the values assumed by the random variable in repeated trials of the experiment. The units on the standard deviation match those of X. Key Takeaways. The probability distribution of a discrete random variable X is a listing of each possible value x taken by X along with the probability P ( x ) that X takes that value in one trial of the experiment.
The mean μ of a discrete random variable X is a number that indicates the average value of X over numerous trials of the experiment. It is computed using the formula μ = Σ x P ( x ). The variance σ 2 and standard deviation σ of a discrete random variable X are numbers that indicate the variability of X over numerous trials of the experiment. They may be computed using the formula σ 2 = Σ x 2 P ( x ) − μ 2, taking the square root to obtain σ. If each die in a pair is “loaded” so that one comes up half as often as it should, six comes up half again as often as it should, and the probabilities of the other faces are unaltered, then the probability distribution for the sum X of the number of dots on the top faces when the two are rolled is x 2 3 4 5 6 7 P ( x ) 1 144 4 144 8 144 12 144 16 144 22 144 x 8 9 10 11 12 P ( x ) 24 144 20 144 16 144 12 144 9 144Compute each of the following. P ( 5 ≤ X ≤ 9 ).
P( X ≥ 7). The mean μ of X.
(For fair dice this number is 7.). The standard deviation σ of X. (For fair dice this number is about 2.415.)Applications. In a hamster breeder's experience the number X of live pups in a litter of a female not over twelve months in age who has not borne a litter in the past six weeks has the probability distribution x 3 4 5 6 7 8 9 P ( x ) 0.04 0.10 0.26 0.31 0.22 0.05 0.02. Find the probability that the next litter will produce five to seven live pups. Find the probability that the next litter will produce at least six live pups. Compute the mean and standard deviation of X.
Interpret the mean in the context of the problem. The number X of days in the summer months that a construction crew cannot work because of the weather has the probability distribution x 6 7 8 9 10 P ( x ) 0.03 0.08 0.15 0.20 0.19 x 11 12 13 14 P ( x ) 0.16 0.10 0.07 0.02. Find the probability that no more than ten days will be lost next summer. Find the probability that from 8 to 12 days will be lost next summer. Find the probability that no days at all will be lost next summer.
Compute the mean and standard deviation of X. Interpret the mean in the context of the problem. An insurance company estimates that the probability that an individual in a particular risk group will survive one year is 0.9825. Such a person wishes to buy a $150,000 one-year term life insurance policy. Let C denote how much the insurance company charges such a person for such a policy. Construct the probability distribution of X. (Two entries in the table will contain C.).
Compute the expected value E ( X ) of X. Determine the value C must have in order for the company to break even on all such policies (that is, to average a net gain of zero per policy on such policies).
Variance Of A Discrete Random Variable Calculator
Determine the value C must have in order for the company to average a net gain of $250 per policy on all such policies. An insurance company estimates that the probability that an individual in a particular risk group will survive one year is 0.99. Such a person wishes to buy a $75,000 one-year term life insurance policy. Let C denote how much the insurance company charges such a person for such a policy. Construct the probability distribution of X. (Two entries in the table will contain C.). Compute the expected value E ( X ) of X.
Determine the value C must have in order for the company to break even on all such policies (that is, to average a net gain of zero per policy on such policies). Determine the value C must have in order for the company to average a net gain of $150 per policy on all such policies. A roulette wheel has 38 slots.
Thirty-six slots are numbered from 1 to 36; half of them are red and half are black. The remaining two slots are numbered 0 and 00 and are green. In a $1 bet on red, the bettor pays $1 to play. If the ball lands in a red slot, he receives back the dollar he bet plus an additional dollar. If the ball does not land on red he loses his dollar. Let X denote the net gain to the bettor on one play of the game.
Construct the probability distribution of X. Compute the expected value E ( X ) of X, and interpret its meaning in the context of the problem. Compute the standard deviation of X. A roulette wheel has 38 slots. Thirty-six slots are numbered from 1 to 36; the remaining two slots are numbered 0 and 00.
Suppose the “number” 00 is considered not to be even, but the number 0 is still even. In a $1 bet on even, the bettor pays $1 to play. If the ball lands in an even numbered slot, he receives back the dollar he bet plus an additional dollar. If the ball does not land on an even numbered slot, he loses his dollar. Let X denote the net gain to the bettor on one play of the game. Construct the probability distribution of X. Compute the expected value E ( X ) of X, and explain why this game is not offered in a casino (where 0 is not considered even).
Compute the standard deviation of X. Haynespro workshopdata torrent. The time, to the nearest whole minute, that a city bus takes to go from one end of its route to the other has the probability distribution shown. As sometimes happens with probabilities computed as empirical relative frequencies, probabilities in the table add up only to a value other than 1.00 because of round-off error. X 42 43 44 45 46 47 P ( x ) 0.10 0.23 0.34 0.25 0.05 0.02.
Find the average time the bus takes to drive the length of its route. Find the standard deviation of the length of time the bus takes to drive the length of its route. Shylock enters a local branch bank at 4:30 p.m. Every payday, at which time there are always two tellers on duty. The number X of customers in the bank who are either at a teller window or are waiting in a single line for the next available teller has the following probability distribution.
X 0 1 2 3 P ( x ) 0.135 0.192 0.284 0.230 x 4 5 6 P ( x ) 0.103 0.051 0.005. What number of customers does Shylock most often see in the bank the moment he enters?.
What number of customers waiting in line does Shylock most often see the moment he enters?. What is the average number of customers who are waiting in line the moment Shylock enters?. The owner of a proposed outdoor theater must decide whether to include a cover that will allow shows to be performed in all weather conditions. Based on projected audience sizes and weather conditions, the probability distribution for the revenue X per night if the cover is not installed is Weather x P ( x ) Clear $3000 0.61 Threatening $2800 0.17 Light rain $1975 0.11 Show-cancelling rain $0 0.11The additional cost of the cover is $410,000. The owner will have it built if this cost can be recovered from the increased revenue the cover affords in the first ten 90-night seasons.
Compute the mean revenue per night if the cover is not installed. Use the answer to (a) to compute the projected total revenue per 90-night season if the cover is not installed. Compute the projected total revenue per season when the cover is in place. To do so assume that if the cover were in place the revenue each night of the season would be the same as the revenue on a clear night. Using the answers to (b) and (c), decide whether or not the additional cost of the installation of the cover will be recovered from the increased revenue over the first ten years. Will the owner have the cover installed?