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Mamont248 [21]
3 years ago
11

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 100 and a standard deviation sigma equa

ls 20. Find the probability that a randomly selected adult has an IQ less than 132. The probability that a randomly selected adult has an IQ less than 132 is?
Mathematics
2 answers:
fomenos3 years ago
5 0

Step-by-step explanation:

First find the z-score.

z = (x - μ) / σ

z = (132 - 100) / 20

z = 1.6

From a z-score table:

P(z<1.6) = 0.9452

So there's a 94.52% probability that a randomly selected adult will have an IQ less than 132.

stealth61 [152]3 years ago
5 0

Answer:

There is a 94.52% probability that a randomly selected adult has an IQ less than 132.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 20. This means that \mu = 100, \sigma = 20.

The probability that a randomly selected adult has an IQ less than 132 is?

This probability is the pvalue of Z when X = 132. So:

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

Z = \frac{132 - 100}{20}

Z = 1.6

Z = 1.6 has a pvalue of 0.9452.

This means that there is a 94.52% probability that a randomly selected adult has an IQ less than 132.

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Answer:

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Step-by-step explanation:

From the question we are told that

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   The probability of getting three or more heads in a row is  P(H) =  0.7870

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Generally  the probability of getting three or more heads in a row and three or more tails in a row is mathematically represented as

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=>   P( H \  n \  T ) =  0.6049

 

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