Question

which of the following probabilities is the greatest for a standard normal distribution?
○ ( p(-1.5leq zleq -0.5) )
○ ( p(-0.5leq zleq 0.5) )
○ ( p(0.5leq zleq 1.5) )
○ ( p(1.5leq zleq 2.5) )

Explanation:

Step1: Use the standard normal distribution property

The standard normal distribution \(Z\sim N(0,1)\) is symmetric about \(z = 0\). Also, the probability \(P(a\leq z\leq b)=\varPhi(b)-\varPhi(a)\), where \(\varPhi(z)\) is the cumulative - distribution function of the standard normal distribution.

We know that \(\varPhi(-z)=1 - \varPhi(z)\)

For \(P(-1.5\leq z\leq - 0.5)=\varPhi(-0.5)-\varPhi(-1.5)=(1 - \varPhi(0.5))-(1 - \varPhi(1.5))=\varPhi(1.5)-\varPhi(0.5)\)

For \(P(-0.5\leq z\leq0.5)=\varPhi(0.5)-\varPhi(-0.5)=\varPhi(0.5)-(1 - \varPhi(0.5)) = 2\varPhi(0.5)-1\)

For \(P(0.5\leq z\leq1.5)=\varPhi(1.5)-\varPhi(0.5)\)

For \(P(1.5\leq z\leq2.5)=\varPhi(2.5)-\varPhi(1.5)\)

Step2: Use the standard - normal table values

From the standard - normal table: \(\varPhi(0.5)\approx0.6915\), \(\varPhi(1.5)\approx0.9332\), \(\varPhi(2.5)\approx0.9938\)

• \(P(-1.5\leq z\leq - 0.5)=\varPhi(1.5)-\varPhi(0.5)=0.9332 - 0.6915=0.2417\)
• \(P(-0.5\leq z\leq0.5)=2\varPhi(0.5)-1=2\times0.6915 - 1=0.383\)
• \(P(0.5\leq z\leq1.5)=\varPhi(1.5)-\varPhi(0.5)=0.2417\)
• \(P(1.5\leq z\leq2.5)=\varPhi(2.5)-\varPhi(1.5)=0.9938 - 0.9332=0.0606\)

Answer:

\(P(-0.5\leq z\leq0.5)\)