Let T represent the fifth test score. The average is the sum of scores divided by their number.
... 82 = (83 +75 +93 +67 +T)/5
... 410 = 318 +T . . . . . . . multiply by 5
... 92 = T . . . . . . . . . . . . subtract 318
The fifth score was 92.
_____
One of the ways I like to work problems like these is to add the differences from the mean. Those are ...
1, -7, 11, -15
so their sum is -10.
The final score must be 10 above the mean in order for this total to be zero. 10 above the mean is 92.
0.84(because 20% of the observations are about 0.84)
You've found the 80th percentile! The 80th percentile of the standard normal distribution is 0.84. That's because 80% of the observations fall below 0.84. (Note: The 80th percentile of every normal distribution is 0.84 times the standard deviation above the mean.)
Answer:
0 is rational
5 rational so u are correct
Step-by-step explanation:
Answer:
$600...
↓↓↓↓↓
Step-by-step explanation:
25x 24 = 600
