Suppose that for a particular computer salesperson, the probability distribution of x = the number of systems sold in 1month is given by the following table.Xp(x)(c)10.03 0.08(b) Find the variance of x.19.7223X45(a) Find the mean value of x (the mean number of systems sold).4.246780.14 0.30 0.30 0.13 0.01 0.01Find the standard deviation of x. (Round your answer to four decimal places.)1.7424How would you interpret these values?The variance of x is the mean squared deviationof the number of systems sold in a month from the meannumber of systems sold in a month. The standard deviation of x is the typical deviationof the numberof systems sold in a month from the mean number of systems sold in a month.What is the probability that the number of systems sold is within 1 standard deviation of its mean value?0.74(d) What is the probability that the number of systems sold is more than 2 standard deviations from the mean?

a) The random variable x defines the number of systems sold in 1 month.The probability distribution of x is, x p(x) 1 0.03 2 0.08 3 0.14 4 0.3 5 0.3 6 0.13 7 0.01 8 0.01 Consider that below table. x p(x) xp(x) x2p(x) 1 0.03 0.03 0.03 2 0.08 0.16 0.32 3 0.14 0.42 1.26 4 0.3 1.2 4.8 5 0.3 1.5 7.5 6 0.13 0.78 4.68 7 0.01 0.07 0.49 8 0.01 0.08 0.64 Total 1 4.24 19.72 The expected value is µ = Σxp(x) = 4.24. The mean value of x is 4.24. b) The variance is, The variance of x is 1.7424. The standard deviation is, The standard deviation of x is 1.3200. The variance of x is the mean squared deviation of the number of systems sold in a month from the mean number of systems sold in a month. The standard deviation of x is the typical deviation of the number of systems sold in a month from the mean number of systems sold in a month. c) Now, µ + σ = 4.24 + 1.32 = 5.56 and µ – σ = 4.24 – 1.32 = 2.92. The probability that number of systems sold is within 1 standard deviation of its mean value is, P(µ – σ < x < µ + σ) = P(2.92 < x < 5.56) = P(x = 3) + P(x = 4) + P(x = 5) = 0.14 + 0.30 + 0.30 = 0.74. Thus, the probability that number of systems sold is within 1 standard deviation of its mean value is 0.74. d) Now, µ + 2σ = 4.24 + (2)(1.32) = 6.88 and µ – 2σ = 4.24 – (2)(1.32) = 1.6. The probability that number of systems sold is more than 2 standard deviations from the mean is, P(x < μ − 2σ or x > μ + 2σ) = 1 – P(µ – 2σ < x < µ + 2σ) = 1 – P(1.6 < x < 6.88) = 1 – [P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6)] = 1 – [0.08 + 0.14 + 0.30 + 0.30 + 0.30 + 0.13] = 1 – 0.95 = 0.05. Thus, the probability that number of systems sold is more than 2 standard deviations from the mean is 0.05.