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Given that x is a normal variable with mean μ = 47 and standard deviation σ = 6.3, find the following probabilities. (Round your answers to four decimal places.) (a) P(x ≤ 60) (b) P(x ≥ 50) (c) P(50 ≤ x ≤ 60)

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Answer:

(a)
P\left ( x\leq 60 \right )=0.9803, (b)
P\left ( x\geq 50 \right )=0.3156, (c)
P(50 \leq x \leq 60)=0.2959

Explanation:

Given that mean
\mu =47 and standard deviation
\sigma =6.3

(a) We need to find
P\left ( x\leq 60 \right ). Using z score


z=(x-\mu )/(\sigma )=(60-47)/(6.3)=2.063492

So,


P\left ( x\leq 60 \right )=P\left ( z\leq 2.063492 \right )=0.9803

(b) We need to find
P\left ( x\geq 50 \right ). Using z score


z=(x-\mu )/(\sigma )=(50-47)/(6.3)=0.4761904

So,


P\left ( x\geq 50 \right )=P\left ( z\geq 0.4761904 ) \right )=1-P\left ( z\leq 0.4761904 ) \right )=1-0.6844=0.3156

(c) We need to find
P(50 \leq x \leq 60). From above z scores


P(50 \leq x \leq 60)=P\left ( 0.4761904\leq z\leq2.063492 \right )=0.2959

User Moaz  Mabrok
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