Suppose I want to construct a confidence interval on the mean run time of a bit of optimization code that I have written. Suppose we know that the run time is approximately normal with variance 225 minutes. I run the code 20 times and find the sample mean of the run times is 325 minutes. If I were to construct a 99% confidence interval of the form

L <= µ <= U

What would be the value of L?

a) The 99% of Control limits are

316.36 < µ< 333.64

b) Lower limit  :  = 316.36

Step-by-step explanation:

Explanation:-

Step(i)

Given the sample size 'n' = 20

Given the sample mean 'x⁻' = 325minutes

Suppose we know that the run time is approximately normal with variance 225 minutes

Population variance 'σ²' = 225 minutes

Standard deviation of the population 'σ' = √225 = 15

Level of significance ∝ = 0.99

Step(ii):-

99% of confidence intervals of the mean is determined by

The z-score of 99% of confidence intervals =2.576

on calculation, we get

(325 - 8.640 , 325 + 8.640)

99% of Control limits are

316.36 < µ< 333.64

Lower limit  : 325 - 8.640 = 316.36

upper limit  : 325 + 8.640 = 333.64

Conclusion:-

a) The 99% of Control limits are

316.36 < µ< 333.64

b) Lower limit  :  = 316.36

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

i can you with this sweetie!

step-by-step explanation: