The grades of an exam whose mean is 82 and whose standard deviation is 14 are normally distributed. Anyone who scores below 55 will be retested. What percentage does this represent

This represents 2.6803%

Step-by-step explanation:

We start by calculating the z-score

Mathematically;

z-score = (x - mean)/SD

here our x is 55

mean = 82

SD = 14

z-score = (55-82)/14

z-score = -1.93

So the probability that a test taker will be retested will be;

P( x < -1.93)

We use the standard normal distribution table for this

P( x < -1.93) = 0.026803

If we convert this to percentage, we have 2.6803%

answer: the answer is 23

step-by-step explanation:

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %