Please provide stepsSolutionGiven that X1X2 and X3 are indep

Please provide steps

Given that X1,X2 and X3 are independent normal random variables with mean common µ1 = 60 and common variance 1² =12.

Given that Y1,Y2 and Y3 are independent normal random variables with mean common µ2 = 65 and common variance 2² =15.

Also Xi and Yj are independent for all i and j.

The distribution of X1 + X2 + X3 is also Normal with mean = 60+60+60=180 and variance = 12+12+12 = 36

X1 + X2 + X3 ~ N ( mean = 180 and variance = 36)

Let X= X1 + X2 + X3

P(X > 185) = P( (X - mean) / sd > (185 - mean)/ sd )

where sd is the square root of variance.

sd = sqrt(36 ) = 6

P( Z > (185 - 180) / 6 ) = P(Z > 0.83) = 1 - P(Z <= 0.83) = 1 - 0.796731 = 0.203269

The distribution of Ybar - Xbar.

The distribution of Ybar is Normal with mean is 65 and variance is 5.

The distribution of Xbar is normal with mean is 60 and variance is 4.

The distribution of Ybar - Xbar is also Normal with mean is 65 - 60 = 5 and variance is 5 + 4 = 9.

Ybar - Xbar ~ N(mean = 5, variance = 9)

Let Ybar - Xbar = X

P(X > 8) = P( (X - mean) / sd > (8-mean)/sd )

= P ( Z > ( 8 - 5 ) / sqrt(9) )

= P ( Z > 3 / 3) = P ( Z > 1) = 1 - P(Z <=1) = 1 - 0.841345 = 0.158655

Let X1 , X2 and X3 denotes the lives of the batteries in each of the 3 flashlights respectively.

All the random variables are independent normal random variables with means 6 , 7 and 8 and variances 2 , 3 and 4 respectively.

We know that X1+X2+X3 follows Normal with mean (6+7+8) = 21 and variance is 2+3+4 = 9

Find the 95th percentile of the total duration of the flashlights.

That means we have given middle area 95% and we have to find the corresponding x.

For 95% the z-score is 1.960.

This we can find by using normal statistical table.

That means 1.960 = (X - mean)/sd

We have 1.960 = (X - 21) / sqrt(9) = (X - 21) / 3

1.960 = (X - 21) / 3

5.88 = X - 21

X = 26.88

P(flashlight will last a total of less than 25 hours) = P( X1 + X2 +X3 < 25)

Let X1+X2+X3 = X

P( X < 25) = P( (X - mean) / sd < ( 25 - mean ) / sd )

= P( Z < (25 - 21) / sqrt(9) ) = P ( Z < 4 / 3) = P(Z < 1.33)

P( X < 25) = 0.908241

Given that in the c) part there are 3 friends have five camping trips that year and each time start with the same types of fresh batteries as above.

P( batteries last more than 25 hours ) = P(X > 25)

P(X > 25) = P( (X- mean /sd) > (25 - mean) / sd )

= P( Z > ( 25 - 21) / sqrt(9) )

= P ( Z > 4 / 3) = P ( Z > 1.33) = 1 - P(Z <= 1.33) = 1 - 0.908241 = 0.091759

Probability that the betteries last more than 25 hours exactly 3 of the 5 times that means we have to multiply the probability by 3.

0.091759 * 3 = 0.275277