Respuesta :
Answer: C. 95%
Step-by-step explanation:
We would apply the Standard Deviation Rule which states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean . The rule is further illustrated below
68% of data falls within the first standard deviation from the mean.
95% fall within two standard deviations.
99.7% fall within three standard deviations.
From the information given, the mean is 100 and the standard deviation is 15. The required IQ score is 130 or more.
2 standard deviations = 2 × 15 = 30
100 + 30 = 130
95% of the IQ scores would fall within two standard deviations.
Therefore, the percent of adults that would qualify for membership is 95%
Answer:
B. 2.5%
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 100, \sigma = 15[/tex]
The organization MENSA, which calls itself 'the high IQ society', requires an IQ score of 130 or higher for membership. About what percent of adults would qualify for membership?
This is 1 subtracted by the pvalue of Z when X = 130. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{130 - 100}{15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% is close to 2.5%
So the correct answer is:
B. 2.5%
