Answer:
Probability of having student's score between 505 and 515 is 0.36
Given that z-scores are rounded to two decimals using Standard Normal Distribution Table
Step-by-step explanation:
As we know from normal distribution: z(x) = (x - Mu)/SD
where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution
Therefore using given data: Mu (Mean) = 510, SD = 10.4 we have z(x) by using z(x) = (x - Mu)/SD as under:
In our case, we have x = 505 & 515
Approach 1 using Standard Normal Distribution Table:
z for x=505: z(505) = (505-510)/10.4 gives us z(505) = -0.48
z for x=515: z(515) = (515-510)/10.4 gives us z(515) = 0.48
Afterwards using Normal Distribution Tables and rounding the values to two decimals we find the probabilities as under:
P(505) using z(505) = 0.32
Similarly we have:
P(515) using z(515) = 0.68
Now we may find the probability of student's score between 505 and 515 using:
P(505 < x < 515) = P(515)-P(505) = 0.68 - 0.32 = 0.36
PS: The standard normal distribution table is being attached for reference.
Approach 2 using Excel or Google Sheets:
P(x) = norm.dist(x,Mean,SD,Commutative)
P(505) = norm.dist(505,510,10.4,1)
P(515) = norm.dist(515,510,10.4,1)
Probability of student's score between 505 and 515= P(515) - P(505) = 0.36
to 4. Your problem should look like: 4 x 1. 
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