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:
For z = 0.525:
A raw score of approximately 82.62 corresponds to the 70th percentile.
Answer:
0.9772 = 97.72% probability that a randomly selected firm will earn more than Arc did last year
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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.
Suppose the mean income of firms in the same industry as Arc for a year is 90 million dollars with a standard deviation of 7 million dollars.
This means that
What is the probability that a randomly selected firm will earn more than Arc did last year?
Arc earned 76 million, so this is 1 subtracted by the pvalue of Z when X = 76.
has a pvalue of 0.0228
1 - 0.0228 = 0.9772
0.9772 = 97.72% probability that a randomly selected firm will earn more than Arc did last year
