The output voltage of a power supply is assumed to be normally distributed. Sixteen observations taken at random on voltage are as follows: 10.35, 9.30, 10.00, 9.96, 11.65, 12.00, 11.25, 9.58, 11.54, 9.95, 10.28, 8.37, 10.44, 9.25, 9.38, and 10.85. I Assume that the standard deviation is known to be 9. Let ¯x be the average of the 16 observations. Find the interval [¯x − a, ¯x + a] such that

CI(lower ==> xbar - a) = 9.769841

CI(upper ==> xbar + a) = 10.748909

Step-by-step explanation:

First, the standard deviation of the sample is not 9 as stated in the question. It is 0.9990493. And the mean of the observation is 10.25938.

The formula for calculation mean and standard deviation are:

Mean = , and

Standard deviation =

By Confident interval (CI),

CI = ,

Where = Standard error (se), and .

If we compute this CI = , we have [9.769841, 10.748909].

For replication of answers, use the R codes below:

d = c(10.35, 9.30, 10.00, 9.96, 11.65, 12.00, 11.25, 9.58, 11.54,  

9.95, 10.28, 8.37, 10.44, 9.25, 9.38, 10.85)

mean(d)

sd(d)

se = sd(d)/sqrt(length(d))

se

CIl = mean(d) - 1.96* se

CIU = mean(d) + 1.96* se

CI = c(CIl, CIU)

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0