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If X is normally distributed with mean 30 and standard deviation 5, the value of P(26≤X≤34)P(26 \le X \le 34) is: (Where P(0≤Z≤0.8)=0.2881P(0 \le Z \le 0.8) = 0.2881)

Solution

✅ Correct Option: 2

Standardising, z=26−305=−0.8z = \frac{26-30}{5} = -0.8 and z=34−305=0.8z = \frac{34-30}{5} = 0.8. By symmetry, P(−0.8≤Z≤0.8)=2(0.2881)=0.5762P(-0.8 \le Z \le 0.8) = 2(0.2881) = 0.5762.

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