Respuesta :
Answer:
The minimum score required for an A grade is 88.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 77.8, \sigma = 8.5[/tex]
Find the minimum score required for an A grade.
Top 12%, which is at least the 100-12 = 88th percentile, which is the value of X when Z has a pvalue of 0.88. So it is X when Z = 1.175.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.175 = \frac{X - 77.8}{8.5}[/tex]
[tex]X - 77.8 = 1.175*8.5[/tex]
[tex]X = 87.8[/tex]
Rounding to the nearest whole number
The minimum score required for an A grade is 88.
Answer: the minimum score required for an A grade is 88.
Step-by-step explanation:
Since the scores on the test are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = scores on the test.
µ = mean score
σ = standard deviation
From the information given,
µ = 77.8
σ = 8.5
The probability value for the top 12% of the scores would be (1 - 12/100) = (1 - 0.13) = 0.88
Looking at the normal distribution table, the z score corresponding to the probability value is 1.18
Therefore,
1.18 = (x - 77.8)/8.5
Cross multiplying by 8.5, it becomes
1.18 × 8.5 = x - 77.8
10.03 = x - 77.8
x = 10.03 + 77.8
x = 88 to the nearest whole number
