The number of dinners served at a restaurant each evening is a random variable w

Question

The number of dinners served at a restaurant each evening is a random variable with mu = 50 and sigma = 5. Using Chebyshev's theorem, at least how many dinners can the restaurant expect to serve each evening, given that (a) k = 2 (b) k = 3? Use Chebyshev's theorem to show that the probability is at least that 35/36 that (a) in 10,000 flips of a balanced coin there will be between 4, 700 and 5, 300 heads, and hence the proportion of heads will be between 0.47 and 0.53; (b) in 1,000,000 flips of a balanced coin there will be between 497,000 and 503,000 heads, and hence the proportion of heads will be between 0.497 and 0.503.

Explanation / Answer

at least 11/k2 of the data lie within k standard deviations of the mean, that is, in the interval with endpoints ± k

6.81 From Chebyshev's theorem

at least 11/k2 of the data lie within k standard deviations of the mean, that is, in the interval with endpoints ± k

Given = 50 = 5 a) k = 2 Endpoints of the data for k= 2 is ± 2 which is 50 ± 2(5) = (40, 60) Thus for k=2, the minimum dinners that the restaurant can expect to serve is 40 b) k = 3 Endpoints of the data for k= 2 is ± 3 which is 50 ± 3(5) = (35, 65) Thus for k=3, the minimum dinners that the restaurant can expect to serve is 35 6.82 a) Since it is a balanced coin, probability of getting a head = p = 0.5 probability of not getting a head = q = 1 - p = 1 - 0.5 = 0.5 Thus it is a binomial distribution with n = 10000,   p = 0.5, q = 0.5 Mean of the binomial distribution = np = 10000 x 0.5 = 5000 Variance of the binomial distribution = npq = 10000 x 0.5 x 0.5 = 2500 Standard Deviation = = 50 = 5000 = 50 For the interval 4700 and 5300 We first distance of the endpoints from the mean 5000 Thus 4700 is (5000-4700 = 300) away from the mean And 5300 is (5300 - 5000 = 300) away from the mean 300 = k = k (50) = 6(50) Thus k = 6 From Chebyshev's theorem 1- 1/k2 of the data lies between k standard deviations from the mean Thus 1 - 1/62 of the heads are between 4700 and 5300 1 - 1/62 = 35/36 Hence P(getting between 4700 and 5300 heads) = 35 / 36 Proportion of heads will be between 4700/10000 and 5300/10000 thus proportion of heads will be between 0.47 and 0.53 b) Since it is a balanced coin, probability of getting a head = p = 0.5 probability of not getting a head = q = 1 - p = 1 - 0.5 = 0.5 Thus it is a binomial distribution with n = 1,000,000,   p = 0.5, q = 0.5 Mean of the binomial distribution = np = 1000000 x 0.5 = 500000 Variance of the binomial distribution = npq = 1000000 x 0.5 x 0.5 = 250000 Standard Deviation = = 500 = 500000 = 500 For the interval 497,000 and 503,000 We first distance of the endpoints from the mean 500000 Thus 497000 is (500000-497000 = 3000) away from the mean And 503000 is (503000 - 500000 = 3000) away from the mean 3000 = k = k (500) = 6(500) Thus k = 6 From Chebyshev's theorem 1- 1/k2 of the data lies between k standard deviations from the mean Thus 1 - 1/62 of the heads are between 497000 and 503000 1 - 1/62 = 35/36 Hence P(getting between 497000 and 503000 heads) = 35 / 36 Proportion of heads will be between 497000/1000000 and 503000/1000000 thus proportion of heads will be between 0.497 and 0.503

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