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Let X denote the number of scores in a test. If X is normally distributed with mean 100 and standard deviation 15, then the probability that X does not exceed 130 is,

(Given P(0≤Z≤2)=0.4772P(0 \le Z \le 2)=0.4772, where ZZ be the standard normal variate)

Solution

✅ Correct Option: 3

Standardise: Z=130−10015=2Z = \dfrac{130-100}{15} = 2.

P(X≤130)=P(Z≤2)=0.5+P(0≤Z≤2)=0.5+0.4772=0.9772P(X\le130) = P(Z\le2) = 0.5 + P(0\le Z\le2) = 0.5 + 0.4772 = 0.9772.

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