Answer:
63%
Step-by-step explanation:
Given the following :
Average cholesterol level or mean (m) = 194
Standard deviation (sd) = 15
what percentage of children have a cholesterol level lower than 199?
P(X < 199)
Assume a normal distribution :
Find the z-score :
Z = (score - mean) / standard deviation
Score = 199
Z = (199 - 194) / 15
Z = 5 / 15
Z - score = 0.3333
P(Z < 0.33) :
Using the z - table ; 0.33 = 0.6293
P(Z < 0.33) = 0.6293
0.6293 * 100% = 62.93%
= 63% (nearest whole percent)