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PEA 2023Question 19descriptiveModerate Under review
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F(1.4)=0.92,  F(0.14)=0.555,  F(−0.2)=0.42,  F(−1.6)=0.055F(1.4)=0.92,\;F(0.14)=0.555,\;F(-0.2)=0.42,\;F(-1.6)=0.055. Diameter ∼N(μ,σ2)\sim N(\mu,\sigma^2). P(diam<1.8)=0.08P(\text{diam}<1.8)=0.08 and P(diam>2.4)=0.055P(\text{diam}>2.4)=0.055. Find μ\mu.

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Final answer

Option D