A binomial distribution consists of 120 trials, with a success probability of 43% . determine the mean and standard deviation
mean μ=np\mu = npμ=np
μ=120×0.43=51.6\mu = 120 \times 0.43 = 51.6μ=120×0.43=51.6
standard deviation= σ\sigmaσ
σ2=np(1−p)=120×0.43(1−0.43)\sigma^2= np(1-p)= 120\times 0.43(1-0.43)σ2=np(1−p)=120×0.43(1−0.43)
σ2=29.412\sigma^2= 29.412σ2=29.412
σ\sigmaσ = 29.412=5.423\sqrt{29.412}= 5.42329.412=5.423
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