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Margarita [4]
3 years ago
12

A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ t

est, what is the probability that the sample mean scores will be between 87 and 124 points
Mathematics
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

100% probability that the sample mean scores will be between 87 and 124 points

Step-by-step explanation:

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

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 105, \sigma = 20, n = 20, s = \frac{20}{\sqrt{20}} = 4.47

What is the probability that the sample mean scores will be between 87 and 124 points

This is the pvalue of Z when X = 124 subtracted by the pvalue of Z when X = 87. So

X = 124

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

By the Central Limit Theorem

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

Z = \frac{124 - 105}{4.47}

Z = 4.25

Z = 4.25 has a pvalue of 1

X = 87

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

Z = \frac{87 - 105}{4.47}

Z = -4.25

Z = -4.25 has a pvalue of 0

1 - 0 = 1

100% probability that the sample mean scores will be between 87 and 124 points

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Step-by-step explanation:

a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

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Looking at the normal distribution table, the probability corresponding to the area above the z score is 1 - 0.99461 = 0.00539

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The difference between sample sample mean and population mean is 9.9 - 9 = 0.9

Since the curve is symmetrical and it is a two tailed test, the x value for the left tail is 9 - 0.9 = 8.1

the x value for the right tail is 9 + 0.9 = 9.9

These means are higher and lower than the null mean. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area. We already got the area above the z score as z = 0.00539

We would double this area to include the area in the left tail of z = - 2.55. Thus

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But we are told to use the critical value approach, then

Since α = 0.01, the critical value is determined from the normal distribution table.

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The z score for an area to the right of 0.995 is 2.575

In order to reject the null hypothesis, the test statistic must be smaller than - 2.575 or greater than 2.575

Since - 2.55 > - 2.575 and 2.55 < 2.575, we would reject the null hypothesis. This corresponds to our previous decision.

b) we would assume normal distribution because the sample size is sufficiently large and the population standard deviation is known.

c) Confidence interval is written in the form,

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Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.

Margin of error = z × σ/√n

The z score for confidence level of 99% is 2.58

Margin of error = 2.58 × 2.5/√50 = 0.91

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The lower end of the confidence interval is

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