Answer:
2.28% probability that a person selected at random will have an IQ of 110 or greater
Step-by-step explanation:
Problems of normally distributed samples are 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.
In this problem, we have that:

What is the probability that a person selected at random will have an IQ of 110 or greater?
This is 1 subtracted by the pvalue of Z when X = 110. So



has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that a person selected at random will have an IQ of 110 or greater
X = -4/3
Hope this helps ,
Give thanks !
I believe it's A, C and D.
Answer:25.5m^3
Step-by-step explanation:
radius(r)=2.3m
π=3.14
volume of hemisphere=2/3 x π x r^3
Volume=2/3 x 3.14x(2.3)^3
Volume=2/3 x 3.14 x 2.3 x 2.3 x 2.3
volume=(2x3.14x2.3x2.3x2.3) ➗ 3
Volume=76.40876 ➗ 3
Volume=25.5