The data below shows the grams of fat for a variety of snacks. Morris wants to calculate the standard error of the sample mean for this set of data. Snack Grams of Fat Snack 1 9 Snack 2 13 Snack 3 21 Snack 4 30 Snack 5 31 Snack 6 31 Snack 7 34 Snack 8 25 Snack 9 28 Snack 10 20 What is the standard error for this set of data

the standard error of data set is 2.628

Step-by-step explanation:

Given the data in the question;

set;

x = 9, 13, 21, 30, 31, 31, 34, 25, 28, 20

To get the standard of Error for this data set, we use the formual

S.E = s / √n

First we determine the mean average;

Mean x' = ∑x / n =  ( 9 + 13 + 21 + 30 + 31 + 31 + 34 + 25 + 28 + 20 ) / 10

x' = 242 / 10

Mean x' = 24.2

Next we find the standard deviation s:

x             (x-x')²

9           231.04

13          125.44

21          10.24

30         33.64

31         46.24

31         46.24

34        96.04

25        0.64

28        14.44

20        17.64

Total ∑(x-x')² = 621.6

so Variance = ∑(x-x')² / (n-1) = 621.6 / ( 10 - 1 )  = 621.6 / 9

Variance = 69.0667

Standard deviation S = √Variance

Standard deviation S = √69.0667

Standard deviation S = 8.3106

So we substitute into our formula to get the standard of error;

S.E = 8.3106 / √10

S.E = 2.628

Therefore, the standard error of data set is 2.628

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

she will most likely have 33  

step-by-step explanation: