You will get nowhere without first solving for the mean, which equals 100: the aggregate of the times (1000) divided by the number of children (10).

You are looking to find the number of children with times that are greater than one standard deviation below (that is, further from) the mean, which you have just determined to be 100. 100 less 22.4 is 77.6. This would have to mean that two of the times, 70 and 75, are even further from the mean (and lower) than 77.6.

Don't let the concept of standard deviation distract you unnecessarily. Furthermore, with single values available in every answer choice, you can disregard the trivial detail of 22.4 having been rounded to the nearest tenth.

70, 75, 80, 85, 90, 105, 105, 130, 130, 130

The list shown consists of the times, in seconds, that it took each of 10 schoolchildren to run a distanc eof 400 meters. If the standard deviation of the 10 running times is 22.4 seconds, rounded to the nearest tenth of a second, how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times.

o One
o Two
o Three
o Four
o Five

First find out mean
Mean = (130*3) + (105*2) + 90+85+80+75+70
= 390+450+160
=1000/10
= 100

1 SD below the mean = 100-22.4 = 77.6
there are only 2 values below this 70 & 75