If we were to toss a single die (singular for dice) 1000 times, how many 6s could we expect over the long run? What is the standard deviation?

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peterpan7117i · Accepted Answer

μ=5003,σ=2532,z=22. Explanation: Let X denote the number of 6&#039;s thrown in 1000 trials. Then X has a binomial distribution as it satisfies the 4 conditions for a random variable to have a binomial distribution. There is a fixed number of trials. (1000) The probability of the event occurring is constant. (1/6) Each throw of the die is independent. There are only two outcomes. (a 6 is thrown or it is not) In general if we had a random variable Z distributed binomially we would write Z∼B(n,p) where n is the number of trials and p is the probability of the event occurring. So we have X∼B(1000,16). We are asked to find the expectation of X and the standard deviation. It can be shown that the expectation of the general binomial random variable Z is np and its variance is np(1-p). See a proof here. So, E(X)=1000⋅16, E(X)=5003, E(X)≈166.67 to 2 decimal places. and Var(X)=1000⋅16⋅56, Var(X)=12509. As σ=Var(X), σ=2532, σ≈11.79 to 2 decimal places. The z score is the number of standard deviations away from the mean. A result of 200 gives a z score of z=200−50032532 z=(10032532), z=42 z=22 z=2.83 to 2 decimal places.

amarantha41 · Accepted Answer

Let X = resulting number of faces as six in rolling a die. Now, p=Probability (X=6)=16 For n =1000 The expected number of getting a number six is: E(X)=1000⋅16≈166.667 The standard deviation for the number of six in long run is: np(1−p)=1000⋅16⋅(1−16)≈11.7851 z=200−50032532 z=(10032532), z=42 z=22 z=2.83

If we were to toss a single die (singular for

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