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Im not sure but i think the answer is the first one y/x of point b= y/x of point A
hope this helps
Answer: 0.86 of the exam scores are between 68 and 77.99 points
Step-by-step explanation:
Since the set of computer science exam scores are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = computer science exam scores .
µ = mean score
σ = standard deviation
From the information given,
µ = 71.33 points
σ = 3 points
We want to find the proportion of the exam scores are between 68 and 77.99 points. It is expressed as
P(68 ≤ x ≤ 77.99)
For x = 68,
z = (68 - 71.33)/3 = - 1.11
Looking at the normal distribution table, the probability corresponding to the z score is 0.13
For x = 68,
z = (77.99 - 71.33)/3 = 2.22
Looking at the normal distribution table, the probability corresponding to the z score is 0.99
P(68 ≤ x ≤ 77.99) = 0.99 - 0.13 = 0.86
