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
Answer:
The nth term is 109-9n
Step-by-step explanation:
Here, we want to find the nth term of the given arithmetic sequence
Mathematically, we have the nth term as;
Tn = a + (n-1)d
where a is the first term which is 100 in this case
d is the common difference which is the value obtained by subtracting the preceding term from the succeeding term; it is constant throughout the sequence
The value here is thus;
82-91 = 91-100 = -9
Substituting these values
Tn = 100 + (n-1)-9
Tn = 100 -9n + 9
Tn = 100 + 9 - 9n
Tn = 109-9n
Find the attachments for solution and explanation
Answer:
1 1/4
Step-by-step explanation:
x=2, y=3, and z=1
1/2(2)+3/4(3)=1+3/12=1 1/4
Answer:
Easy it's a right hand child
Step-by-step explanation:
Because if both husband and wife both have right hands. And it's 10% chance the child will be left handed. Then it's possible that the child is going to be a right hand.
