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katovenus [111]
3 years ago
11

some psychologists believe that a genius should be defined as anyone having an IQ over 140. If IQ scores are normally distribute

d with a mean of 100 and a standard deviation of 17 and if the population of the world is 6,575,000,000, how many geniuses are there in the world today
Mathematics
2 answers:
denis23 [38]3 years ago
6 0

Answer:

61,239,550

Step-by-step explanation:

We let the random variable X denote the IQ scores. This would imply that X is normal with a mean of 100 and standard deviation of 17. We proceed to determine the probability that an individual chosen at random from the population would be a genius, that is;

Pr( X>140)

The next step is to evaluate the z-score associated with the IQ score of 140 by standardizing the random variable X;

Pr(X>140)=Pr(Z>\frac{140-100}{17})=Pr(Z>2.3529)

The area to the right of 2.3529 will be the required probability. This area from the standard normal tables is 0.009314

From a population of 6,575,000,000 the number of geniuses would be;

6,575,000,000*0.009314 = 61,239,550

andreyandreev [35.5K]3 years ago
4 0

Answer:

61,805,000

Step-by-step explanation:

lets assume, no of geniuses = X

P(X > 140) = P(z > (140 - 100)/17)

                = P(z > 40/17) ⇒ P(z > 2.352)

1 - P(z < 2.352) = 1 - 0.9906 ⇒ 0.0094

The number of geniuses in the world today are:

0.0094 x 6,575,000,000 = 61,805,000

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Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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Problems of normal distributions can be solved using the z-score formula.

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The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

(a) If the true mean is 0.9370 with a standard deviation of 0.0090 within what interval will 99% of the sample means fail?

Samples of 34 means that n = 34

We have that \mu = 0.937, \sigma = 0.009

By the Central Limit Theorem, s = \frac{0.009}{\sqrt{34}} = 0.0015

Within what interval will 99% of the sample means fail?

Between the (100-99)/2 = 0.5th percentile and the (100+99)/2 = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

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

By the Central Limit Theorem

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

-2.575 = \frac{X - 0.937}{0.0015}

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99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

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

2.575 = \frac{X - 0.937}{0.0015}

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99% of the sample means will fall between 0.933 and 0.941.

(b) If the true mean 0.9370 with a standard deviation of 0.0090, what is the sampling distribution of ¯X?

By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

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