What is 531 ÷ 16 and the remander
2 answers:
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4y^2 + 2 - 2 = 38 - 2 (subtract both side by 2)
4y^2 = 36
Divide by 4:
y^2 = 9
y =
y = 3
Answer:
The cutoff value for the top 5% of all IQs is 131.255.
Step-by-step explanation:
When the distribution is normal, we use 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.
In this question, we have that:
What is the cutoff value for the top 5% of all IQs
This is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.
The cutoff value for the top 5% of all IQs is 131.255.