Answer :
Answer:
82.62
Step-by-step explanation:
Mean score (μ) = 80
Standard deviation (σ) = 5
The 70th percentile of a normal distribution has an equivalent z-score of roughly 0.525.
For any given score, X, the z-score can be determined by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For z = 0.525:
[tex]0.525=\frac{X-80}{5}\\ X=82.62[/tex]
A raw score of approximately 82.62 corresponds to the 70th percentile.
Answer: the raw score that corresponds to the 70th percentile is 82.625
Step-by-step explanation:
Since the population of scores in the aptitude test is normally distributed., we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = aptitude test scores.
µ = mean score
σ = standard deviation
From the information given,
µ = 80
σ = 5
We want to find the raw score that corresponds to the 70th percentile.
70th percentile = 70/100 = 0.7
Looking at the normal distribution table, the z score corresponding to 0.7 is 0.525.
Therefore,
0.525 = (x - 80)/5
5 × 0.525 = x - 80
2.625 = x - 80
x = 2.625 + 80
x = 82.625