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)

i would say c is the best answer, hope it's right!

good luck! =)

cutie~

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1