Electricity usage data consists of 45 months has a mean number of units consumed is 390.47 per month with a standard deviation of 170.5 units per month. Assume that the number of units consumed are approximately normally distributed. Estimate 95% confidence interval for the average monthly electricity consumed units.

The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87

Step-by-step explanation:

We have the standard deviation for the sample. So we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 45 - 1 = 44

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0141

The margin of error is:

M = T*s = 2.0141*170.5 = 343.4

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 390.47 - 343.40 = 47.07 units per month

The upper end of the interval is the sample mean added to M. So it is 390.47 + 343.40 = 733.87 units per month

The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87

step-by-step explanation: it is d

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !