Suppose one is rolling a four-sided fair die independently and repeatedly for 10

Question

Suppose one is rolling a four-sided fair die independently and repeatedly for 10 times. Let X be the number of rolls that he gets a one. Using the table provided (1). P(X>=3) and (2). P(2 <=X < 5): Show your work.     If part of the table is missing, tell me where to look as the table continues for 5 more pages.

Table The Poisson Distribution Fx) 1.0 0.20 0.8 0.15 Poisson, = 3.8 0.6 Poisson, a = 3.8 0.10 Fx. 0.4 0.05 0.2 10 12 10 12 1. F(x) = P(X x) k=0 E(X) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.905 0.995 1.000 1,000 1,000 0.819 0.982 0.999 1,000 1,000 0.741 0.963 0.996 1,000 1,000 0.670 0.938 0.992 0.999 1,000 0.607 0.910 0.986 0.998 1,000 0.549 0.878 0.977 0.997 1,000 0.497 0.844 0.966 0.994 0.999 0.449 0.809 0.953 0.991 0.999 0.407 0.772 0.937 0.987 0.998 0.368 0.736 0.920 0.981 0.996 1.000 1.000 1,000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 100 1,000 1.000 1.000 1.000 1.000 1.000 1,000 0.999 1.000 1.1 1.2 1.3 1.5 1.6 1.8 19 2.0 0.333 0.699 0.900 0.301 0.663 0.879 0.966 0.992 0.273 0.627 0.857 0.957 0,989 0.247 0.592 0.833 0.946 0.986 0.223 0.558 0.809 0.202 0.525 0.783 0.921 0.976 0.183 0.493 0.757 0.907 0.970 0.165 0.463 0.731 0.891 0.964 0.150 0.434 0.704 0.875 0.956 0.135 0.406 0.677 0.857 0.947 0.934 0.974 0.995 0.981 0.996 0.992 0.999 0.998 0.999 1.000 1.000 1,000 0.998 1.000 1.000 1,000 0.998 1.000 1.000 1,000 0.997 0.999 1,000 1,000 0.994 0.999 1.000 1,000 0.990 0.997 0.999 1,000 0.987 0.997 0.999 1,000 0.983 0.995 0.999 1.00 1,000 1,000 1,000 1,000 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 0.074 0.267 0.111 0.355 0.623 0.819 0.928 0.091 0.308 0.570 0.779 0.904 0.518 0.736 0.877 0.06 0.231 0.469 0.692 0.848 0.050 0.199 0.423 0.647 0.815 0.04 0.171 0.38 0.603 0.78 0.033 0.147 0.340 0.558 0.744 0.027 0.126 0.303 0.515 0.706 0.022 0.107 0.269 0.473 0.668 0.018 0.092 0.238 0.433 0.629 0.916 0.975 0.993 0.998 1,000 1.000 0.964 0.988 0.997 0.999 1000 0.951 0.983 0,995 0.999 1,000 0.935 0.976 0.992 0.998 0.999 0.966 0.988 0.996 0.895 0.955 0.983 0.994 0.871 0.942 0.977 0.992 0.997 0.844 0.927 0.969 0.988 0.996 0.816 0.909 0.960 0.984 0.785 0.889 0.949 0.979 0.992 0.999 0.998 0.994 10 1.000 1.000 0.999 1.000 1,000 1.000 1.000 1,000 1.000 1.000 1.000 1.000 1,000 1,000 1.000 0.999 1,000 1.000 0.998 0.999 1.000 0.997 0.999 1.000 12 1.000 1000 1000

Explanation / Answer

For, average value of 3.8, you need to see the third table. There on the top row you will see 2.2, 2.3 and so on. In our case we will see the column corresponding to 3.8.

So, F(1) = P(X < 1) = 0.107

1) P(X > 3) = 1 - P(X < 2) = 1- F(2) = 1 - 0.269 = 0.731

2) P(2 < X < 5) = P(X < 4) - P(X < 1) = F(4) - F(1) = 0.668 - 0.107 = 0.561

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