Respuesta :
A grade of 69.77 is required to get a b or higher in the class.
Normal Probability Distribution
The z-score of a measure X of a normal variable that has mean represented by [tex]\mu[/tex] and standard deviation represented by [tex]\sigma[/tex] is given by the rule presented as follows:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure X is above or below the mean, depending if the z-score is positive or negative.
- From the z-score table, the p-value associated with the z-score is found, and it represents the percentile of the measure X.
In the context of this problem, the mean is given as follows:
[tex]\mu = 62[/tex]
The first quartile is of 52, meaning that when X = 52, Z = -0.675, thus we can find the standard deviation as follows:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{52 - 62}{\sigma}[/tex]
[tex]0.675\sigma = 10[/tex]
[tex]\sigma = \frac{10}{0.675}[/tex]
[tex]\sigma = 14.8[/tex]
The top 30% of the scores is composed by scores at the 70th percentile or higher, hence the lower bound is of X when Z = 0.525.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
0.525 = (X - 62)/14.8
X - 62 = 0.525 x 14.8
X = 69.77.
A similar problem, also about the normal distribution, is given at https://brainly.com/question/4079902
#SPJ1
