The data below shows the heights in inches of 10 students in a class. StudentHeight, in inches Student 153 Student 252.5 Student 354 Student 451 Student 550.5 Student 649.5 Student 748 Student 853 Student 952 Student 1050 The standard error of the sample mean for this set of data is . Answer choices are rounded to the hundredths place. 0.59 1.87 1.77 0.19

Question not correct, so i have attached the correct question.

SE = 0.59

Step-by-step explanation:

The mean of the students height is;

x' = (53 + 52.5 + 54 + 51 + 50.5 + 49.5 + 48 + 53 + 52 + 50)/10

x' = 51.35

Now, deviation from the mean for each height;

53 - 51.35 = 1.65

52.5 - 51.35 = 1.15

54 - 51.35 = 2.65

51 - 51.35 = -0.35

50.5 - 51.35 = -0.85

49.5 - 51.35 = -1.85

48 - 51.35 = -3.35

53 - 51.35 = 1.65

52 - 51.35 = 0.65

50 - 51.35 = -1.35

Now, square of the deviations above;

1.65² = 2.7225

1.15² = 1.3225

2.65² = 7.0225

-0.35² = 0.1225

-0.85² = 0.7225

-1.85² = 3.4225

-3.35² = 11.2225

1.65² = 2.7225

0.65² = 0.4225

-1.35² = 1.8225

Sum of the squared deviations;

2.7225 + 1.3225 + 7.0225 + 0.1225 + 0.7225 + 3.4225 + 11.2225 + 2.7225 + 0.4225 + 1.8225 = 31.525

Let's divide the sum by the DF of n - 1 i.e 10 - 1 = 9.

Thus;

31.525/9 = 3.50278

Taking the square root of that gives us the standard deviation.

Thus;

s = √3.50278

s = 1.8716

Formula for standard error is;

SE = s/√n

SE = 1.8716/√10

SE = 0.59

lol.

it's ok we all do this sometimes.

-aparri