3.1.6. Let Y be the number of successes throughout n independent repetitions ofa random experiment with probability of success p = . Determine the smallestvalue of n so that P(1

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Answered: 3.1.6. Let Y be the number of successes…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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3.6.2 Suppose X is a negative binomial random variable with p=0.2, r=4. Find: a. E(X) b. P(X=20) c. Most likely value for X
7.4.15-T Question Help v A humane society claims that less than 73% of households in a certain country own a pet. In a random sample of 600 households in that country, 420 say they own-a pet. At a = 0.10, is there enough evidence to support the society's claim? Complete parts (a) through (c) below. (a) Identify the claim and state Ho and H Identify the claim in this scenario. Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a decimal. Do not round.) O A. More than % of households in the country own a pet. O B. Less than % of households in the country own a pet. O C. The percentage households in the country that own a pet is not %. O D. % of households in the country own a pet. Click to select and enter your answer(s) and then click Check Answer. 4 parts remaining Clear All Check Answer 1:06 AM P Type here to search 2/21/2021 on fs 10 144 esc @ %23 %24 tbacksoace 3. 6. 7. 8. 9. W R Y tab H. K
3. 7 8. 10 An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let R be the random variable counting the number of tails in each outcome. For example, if the outcome is tht, then R (tht)=2. Suppose that the random variable X is defined in terms of R as follows: X=R-1. The values of X are given in the table below. tth hht hth thh ttt hhh htt tht Outcome Value of X 1 - 1 1 1 Calculate the values of the probability distribution function of X, i.e. the function py. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value X of X Px (x) Submit A Continue © 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center étv MacBook Air DII DD F10 80 F9 F7 F8 F6 F4 F5 F2 F3 公云。 olo
9. This is a simplified inventory problem. Suppose that it costs c dollars to stock anitem and that the item sells for s dollars. Suppose that the number of items thatwill be asked for by customers is a random variable with the frequency functionp(k). Find a rule for the number of items that should be stocked in order tomaximize the expected income. (Hint: Consider the difference of successiveterms.)
5.22 a. Form the probability distribution table for P(x) = , for x = 1,, 2, 3. %3D b. Find the extensions xP(x) and x²P(x) for each x. Find E[xP(x)] and E[x*P(x)]. d. Find the mean for P(x) = , for x = 1, 2, 3. 2 e. Find the variance for P(x) =, for x = 1, 2, 3. 6' f. Find the standard deviation for P(x) = *, for o 62 aluvilidsdon х%3D 1, 2, 3.
Q2. Suppose X and S2are the sample mean and sample variance associated with a random sample of size n = 16 from a N (u = 50, o? =81) distribution. Find the value of the constants b and c such that they satisfy the following statement X-50 (a) P스
8. The time (in minutes) that it takes a car mechanic to replace a car battery is a random variable X that follows an exponential distribution with mean 15 minutes. a) Find P(14 < X < 22). Round off your answer to two decimal places. x 22 1 P(14 < X <22) = √₂/1² 15 e 15dx = P(x < 22) — P(x < 14) x 15 122 = e 15 22 1 - e 14 or 0.8 = = 22 14 = −e 15 + e15 = 0.16254≈ 0.16 - (1 – b) Find the 80th percentile. Round off to the nearest whole number. 1 X S 15 14 e 15 t e 15 dt X 0.8 = 1 e 15 X e 15 = = 0.2 14 22 = e 15 e 15 = 0.16 x — = ln (0.2). This gives x = 24.14156 15
6.2/3
7.54 A random variable is Poisson distributed with A = 0.50 arrivals per minute. For the corresponding exponential distribution, and x = minutes until the next arrival, identify the mean of x and determine the following: a. P(x S 0.5) b. P(x S1.5) с. Р\х 2.5) d. P(x 3.0)
11.1.13 Construct a confidence interval for p, - P2 at the given level of confidence. X1 = 375, n, = 505, x2 = 436, n2 = 586, 99% confidence The researchers are % confident the difference between the two population proportions, p1 -P2, is between and (Use ascending order. Type an integer or decimal rounded to three decimal places as needed.)
5.2.1 The conditional probability distribution of Y given X = x is fy.0) = xe-y for y> 0, and the marginal probability distribution of X is a continuous uniform distribution over 0 to 10. a. Graph fyv) = xe¬w for y > 0 for several values of x. Determine: b. P(Y < 2|X = 2) d. E(Y |X = x) f. fr(y) c. E(Y | X = 2) e. fxy(x, y)
M5

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3.1.6. Let Y be the number of successes throughout n independent repetitions of a random experiment with probability of success p = . Determine the smallest value of n so that P(1