A researcher wants to investigate the effects of environmental factors on IQ scores. For an initial study, she takes a sample of 400 people who grew up as the only child. She finds that 48.5% of them have an IQ score over 100. It is known that 50% of the general population has an IQ score exceeding 100.(a) Find the mean of p, where p is the proportion of people with IQ scores over 100 in a random sample of 400 people.(b) Find the standard deviation of p.(c) Compute an approximation for P(p is greater than or equal to 0.485), which is the probability that there will be 48.5% or more individuals with IQ scores over 100 in a random sample of 400. Round answer to 4 decimal places.

Answer:

a) p = 0.5.

b) s = 0.025.

c) 0.7257 = 72.57% probability that there will be 48.5% or more individuals with IQ scores over 100 in a random sample of 400.

Step-by-step explanation:

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

Normal Probability Distribution:

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

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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

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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

It is known that 50% of the general population has an IQ score exceeding 100. Sample of 400.

This means that [tex]n = 400, p = 0.5, s = \sqrt{\frac{0.5*0.5}{400}} = 0.025[/tex]

(a) Find the mean of p, where p is the proportion of people with IQ scores over 100 in a random sample of 400 people

By the Central Limit Theorem, p = 0.5.

(b) Find the standard deviation of p.

By the Central Limit Theorem, s = 0.025.

(c) Compute an approximation for P(p is greater than or equal to 0.485), which is the probability that there will be 48.5% or more individuals with IQ scores over 100 in a random sample of 400.

This is 1 subtracted by the p-value of Z when X = 0.485. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.485 - 0.5}{0.025}[/tex]

[tex]Z = -0.6[/tex]

[tex]Z = -0.6[/tex] has a p-value of 0.2743.

1 - 0.2743 = 0.7257.

0.7257 = 72.57% probability that there will be 48.5% or more individuals with IQ scores over 100 in a random sample of 400.

I don't understand plz help

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