Answer:
Explanation:
Since this distribution is normal,
we can use the z score formula.
z=(x-μ)/σ
Where x is some random value we choose
Where μ is the mean(6)
Where σ is the stdev(1.2)
We need to find values between the times x=0 and x=5, which is considered ideal so let use two z scores.
When x=0:

This is a z score of -5, this is VERY,VERY, unlikely to occur so this percentage can be considered 0%
When x=5:

Using the z score table, we get a percentage about 20.33%
Since the z score gives us a percentage of data is left of the point of interest , we subtract the larger percentage- smaller percentage
20.33%-0= 20.33%= 20%
So about 20% percent of the responses is considered ideal.