A person must score in the upper 8% of the population on an IQ test to qualify for membership in MENSA, the international high-I

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A person must score in the upper 8% of the population on an IQ test to qualify for membership in MENSA, the international high-I

Q society. IQ scores are normally distributed with a mean of 115 and a standard deviation of 17. (a). [7pts] What IQ score must a person get to qualify

Mathematics

2 answers:

nikklg [1K] · Answer 1 · 2022-04-30T09:40:49+03:00

Answer: The person must score an IQ score of 139 to qualify.

Step-by-step explanation:

Since the IQ scores are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = scores on the test.

µ = mean score

σ = standard deviation

From the information given,

µ = 115

σ = 17

The probability value for the upper 8% of the population would be (1 - 8/100) = (1 - 0.08) = 0.92

Looking at the normal distribution table, the z score corresponding to the probability value is 1.41

Therefore,

1.41 = (x - 115)/17

Cross multiplying by 17, it becomes

1.41 × 17 = x - 115

23.97 = x - 115

x = 23.97 + 115

x = 139 to the nearest whole number

Cloud [144] · Answer 2 · 2022-04-30T08:28:19+03:00

Answer:

A person must get an IQ score of at least 138.885 to qualify.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 115, \sigma = 17$