Respuesta :
Using the normal distribution, it is found that David must score at least 109 to pass the exam.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 121, \sigma = 12[/tex]
The threshold is the 16th percentile of scores, which is X when Z = -0.995, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.995 = \frac{X - 121}{12}[/tex]
X - 121 = -0.995(12)
X = 109.
David must score at least 109 to pass the exam.
More can be learned about the normal distribution at https://brainly.com/question/4079902
#SPJ1
