Question

Q1. let x be normal with mean 10 and variance 4. find p(x>12), p(x<10), p(x<11), p(9<x<13)

Q2.Let X be normal with mean 3.6 and variance 0.01. Find C such that P(X<=c)=50%, P(X>c)=10%, P(-c<X-3.6<=c)=99.9%

Answer 1

I will answer the first question. Please add the second question as a separate question.

We know that x is normal.

Mean(x) is 10.

Variance(x) is 4.

Thus, standard deviation(x) is 2.

We must find the z-value (standardized value) of every x value given in the questions.

All probabilities for an associated z value can be seen from a z table.

P(x>12)

P(x>12)= 1-P(x<12)

= 1- Z((12-10)/2)

= 1- Z(1)

= 1-0.8413

= 0.1587 = 15.87%

P(x<10)

= Z((10-10)/2)

= Z(0)

= 0.5000 = 50%

P(x<11)

= Z((11-10)/2)

= Z(0.5)

= 0.6915 = 69.15%

P(9<x<13)

= Z((13-10)/2) - Z((9-10)/2)

= Z(1.5) - Z(-0.5)

= 0.9332 - 0.3085

= 0.6247 = 62.47%