Anormal distribution has a meaning of 50 and a standard deviation of 6. what is the probability that a value selected at random from this data is in the interval from 50 to 62 ? express your answer as a percent rounded to the nearest 10th

We first need to standardise the value X=50 and X=62, in other words, find the z-scores.

The formula to find z-score is (X-μ)/σ

Where μ is the mean score which in this case is 50 and σ is the standard deviation = 6

Z-score for X = 50 ⇒ (50-50)÷6 = 0÷6 = 0 
Z-score for X = 62 ⇒ (62-50)÷6 = 12÷6 = 2

Presenting this information on a normal distribution bell curve, the area we need is in between z = 0 and z = 2

P(0<z<2) = P(z<2) - P(z<0)
P(0<z<2) = 0.9772 - 0.5
P(0<z<2) = 0.4772

As a percentage, the probability is 0.4772 × 100 = 47.72%