Answer:
The score that cuts off the bottom 2.5% is 48.93.
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 score that cuts off the bottom 2.5%
This is X when Z has a pvalue of 0.025, so X when Z = -1.96.




The score that cuts off the bottom 2.5% is 48.93.
The answer is a.
Substitute (-2,1) and (1,10) into each answer. Only a is correct because you will get -7 from both points.
Answer:
x = -33
Step-by-step explanation:
2/3(x+6)=-18
Multiply each side by 3/2
3/2*2/3(x+6)=-18*3/2
x+6 = -27
Subtract 6 from each side
x +6-6 = -27-6
x = -33
Answer:
And using the normal standard table or excel we find the probability:

Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the avergae number of weeks an individual is unemployed of a population, and for this case we know the distribution for X is given by:
Where
and
Since the distribution for X is normal then, the distribution for the sample mean
is given by:
We select a sample of n =50 people. And we want to find the following probability
And using the normal standard table or excel we find the probability:
