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2 answers:
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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:
a)
b)
c)
Step-by-step explanation:
Some previous concepts
The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.
A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".
The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"
The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"
Hypothesis
Null hypothesis:
Alternative hypothesis:
If the random variable is distributed like this:
We assume that the variance is known so the correct test to apply here is the z test to compare means, the statistic is given by the following formula:
Since we have the values for the statistic already calculated we can calculate the p value using the following formulas:
Part a
And in order to find the answer using excel we can use the following code:
"=1-NORM.DIST(2.05,0,1,TRUE)"
Part b
And in order to find the answer using excel we can use the following code:
"=1-NORM.DIST(-1.84,0,1,TRUE)"
Part c
And in order to find the answer using excel we can use the following code:
"=1-NORM.DIST(0.4,0,1,TRUE)"
Conclusions
If we use a reference value for the significance, let's say . For part a the
so then we can reject the null hypothesis at this significance level.
For part b the so then we FAIL to reject the null hypothesis at this significance level.
For part c the so again we FAIL to reject the null hypothesis at this significance level.