Question

in the north racial violence

1. (15pts) Consider the following data: 2 4 5 6 8 P(x) 0.1 0.1 0.3 0.2 0.2 0.1 Step 1: The Expected Value E(X) is Round your answer to one decimal. Step 2: The Variance is Round your answer to at least two decimal places. Step 3: The Standard Deviation is Round your answer to at least two decimal places. Step 4: The value of POX>5) is Round your answer to one decimal. Step 5 The value of P(X s 4) is Round your answer to one decimal. 2. (10 pts) Suppose that you toss a fair coin 40 times. Use the normal distribution and the continuity correction to approximate the probability that more than 18 heads will be observed.

Answer 1

1.) The expected value, E(X)

1.) The expected value, E(X) E(x) = mu_x = Sigma (x_i times P(x_i)) From above given data, rightarrow E(x) = (x_1 times P(x_1)) + (x_2 times P(x_2)) + (x_3 times P(x_3)) + (x_4 times P(x_4)) + (x_5 times P(x_5)) + x_6 times P(x_6)) rightarrow E(x) = (2 times 0.1) + (3 times 0.1) + (4 times 0.3) + (5 times 0.2) + (6 times 0.2) + (8 times 0.1) rightarrow E(x) = (0.2) + (0.3) + (1.2) + (1) + (1.2) + (0.8) = 1.7 variable sigma^2 = E(X^2) - mu^2 sigma^2 = E(X^2) - E(X)^2 rightarrow E(X)^2 = Sigma (x^2_i times P(x_i)) rightarrow E(X^2) = (x_i^2 times P(x_1)) + (x_2^2 times P(X_2)) + (x_3^2 times P(x_3)) + (x_4^2 times P(x_4)) + (x_5^2 times P(x_5)) + (_6^2 times P(x_6)) rightarrow (X^2) = (2^2 times 0.1) + (3^2 times 0.1) + (3^2 times 0.1) + (4^2 times 0.3) + (5^2 times 0.2) +

From above given data,

2.) Variance

$\Rightarrow E(X^2) = ( 0.4) + ( 0.9) + (4.8) + (5) +(7.2) + (6.4 ) = 24.7$

3.) Standard Deviation,

$Standard \,\, deviation = \sqrt{2.61 }$

4.) P(X>5) = ?

P(X> 5) = P(X=6) + P(X=8)

P(X>5) = 0.2 + 0.1 = 0.3

5.) P(X <= 4) = ?

$P(X \leq 4) = P(X=2) + P(X=3) + P(X=4)$

$\Rightarrow P(X \leq 4) = 0.1 + 0.1 + 0.3 = 0.5$

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