Scores on an aptitude test are distributed with a mean of 220 and a standard deviation of 30. The shape of the distribution is unspecified. What is the probability that the sampling error made in estimating the population mean by the mean of a random sample of 50 test scores will be at most 5 points i. e. P(215< <225)?

The probability that the sampling error made in estimating the population mean by the mean of a random sample of 50 test scores will be at most 5 points i.e. P(215< <225) is 0.7620.

Given :

According to the central limit theorem:

Substitute the values of known terms in the above formula.

Simplify the above expression in order to determine the value of 'Z'.

Z = 1.18

The p-value regarding Z = 1.18 is 0.8810.

Now, at X = 215 the central limit theorem becomes:

Simplify the above expression in order to determine the value of 'Z'.

Z = -1.18

The p-value regarding Z = 1.18 is 0.1190.

Now, the probability that the sampling error made in estimating the population mean by the mean of a random sample of 50 test scores will be at most 5 points, that is:

P(215< <225) = 0.8810 - 0.1990

                       = 0.7620

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P(215<X<225) = 0.7620

Step-by-step explanation:

The shape of the distribution is unknow, however, the shape of the sampling distributions of the sample mean is approximately normal due to the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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 normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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

In this problem, we have that:

P(215<X<225)

This is the pvalue of Z when X = 225 subtracted by the pvalue of Z when X = 215. So

X = 225

By the Central Limit Theorem

has a pvalue of 0.8810

X = 215

has a pvalue of 0.1190

0.8810 - 0.1190 = 0.7620

P(215<X<225) = 0.7620

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step-by-step explanation:

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i