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olga nikolaevna [1]
3 years ago
9

On a normal distribution if IQ test scores, with a mean of 100 and a standard deviation of 15 points, a score of 85 places you a

pproximately in what percentile of the population?A. 16thB. 50thC. 97thD. 76thE. 24th
Mathematics
1 answer:
Natali [406]3 years ago
3 0

Answer:

A. 16th

Step-by-step explanation:

We have been given that on a normal distribution IQ test scores, with a mean of 100 and a standard deviation of 15 points. We are asked to find the percentile for a score of 85.

We will use z-score formula and normal distribution table to answer our given problem.

z=\frac{x-\mu}{\sigma}, where,

z = Z-score,

x = Sample score,

\mu = Mean,

\sigma = Standard deviation.

Upon substituting our given values in z-score formula, we will get:

z=\frac{85-100}{15}

z=\frac{-15}{15}

z=-1

Now, we need to find area under a normal curve that corresponds to z-score of -1 that is P(z.

Using normal distribution table, we will get:

P(z

Let us convert 0.15866 into percentage as:

0.15866\times 100\%=15.866\%\approx 16\%

Therefore, a score of 85 places you approximately in 16th percentile of the population.

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Answer:

n=(\frac{2.054(12)}{4})^2 =37.97 \approx 38

So then the minimum sample to ensure the condition given is n= 38

Step-by-step explanation:

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\mu population mean (variable of interest)

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Solution to the problem

When we create a confidence interval for the mean the margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 96% of confidence interval now can be founded using the normal distribution. The significance is \alpha=1-0.96 =0.04. And in excel we can use this formula to find it:"=-NORM.INV(0.02;0;1)", and we got z_{\alpha/2}=2.054, replacing into formula (b) we got:

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So then the minimum sample to ensure the condition given is n= 38

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