The question goes as follows: Four rods A,B,C, and D are to be welded end to end

Question

The question goes as follows: Four rods A,B,C, and D are to be welded end to end to make a45 centimeter rod. The length of each of the rods is a normallydistributed random variable with means µ and standarddeviations given in the table below. What is the probabilitythat the assembled rods will be within 1 centimeter of 45centimeters?
A: =10cm, =1cm B: =11cm, = 2cm C: =12cm, = 2cm D: =13cm, = 4cm
I am not sure how to approach this question Please help! Thank you! The question goes as follows: Four rods A,B,C, and D are to be welded end to end to make a45 centimeter rod. The length of each of the rods is a normallydistributed random variable with means µ and standarddeviations given in the table below. What is the probabilitythat the assembled rods will be within 1 centimeter of 45centimeters?
A: =10cm, =1cm B: =11cm, = 2cm C: =12cm, = 2cm D: =13cm, = 4cm
I am not sure how to approach this question Please help! Thank you!

Explanation / Answer

adding on to my previous reply the sum of a,b,c,d is a gaussianrandom variable thus let the new random variable be x then x will be gaussian with mean= sum of mean of a,b,c,d = 10+11+12+13 = 46 also the variance of x will be the sum f variance of a,b,c,dassuming they are independent of each other hence variance of x = 1+4+4+16 = 5 thus we have to obtain the P{44

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