Scores on an exam follow an approximately Normal distribution with a mean of 74.3 and a standard deviation of 7.4 points. What proportion of students scored below 80 points? Group of answer choices 0.7794 0.2206 0.7703 0.2297 0.8078 0.1922

First option is correct.

Proportion of students with scores less than 80 is 0.7794

Explanation:

We first need to standardize the score 80.

The standardized score for 80 is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ = (80 - 74.3)/7.4 = 0.77

To determine the probability of scoring less than 80, P(x < 80) = P(z < 0.77)

We'll use data from the normal probability table for these probabilities

P(x < 80) = P(z < 0.77) = 1 - P(z ≥ 0.77) = 1 - P(z ≤ -0.77) = 1 - 0.22065 = 0.77935 = 0.7794.

Option A

the answer is b) 2nd

hope this and plz mark as brainliest!

c

explanation: