Q: Calculate the z-score for a test score of 87 if the mean test score is 81.1 and standrd deviation is 11.06?

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If a normally distributed random variable X has mean m and standard deviation s, then z = (X - m)/s

T-score is used when you don't have the population standard deviation and must use the sample standard deviation as a substitute.

Mean μ = 63.3 Standard deviation σ = 3.82 Standard error σ / √ n = 3.82 / √ 19 = 0.8763681 z = (xbar - μ) / (σ / √ n ) z = (61.6-63.3) / 0.876368 z = -1.9398

The absolute value of the standard score becomes smaller.

78

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A z-score cannot help calculate standard deviation. In fact the very point of z-scores is to remove any contribution from the mean or standard deviation.

You need the mean and standard deviation in order to calculate the z-score. Neither are given.

If it is possible to assume normality, simply convert the desired score to a z-score, and look up the probability for that.

score of 92

If a normally distributed random variable X has mean m and standard deviation s, then z = (X - m)/s

T-score is used when you don't have the population standard deviation and must use the sample standard deviation as a substitute.

zero

The standardised score decreases.

When you don't have the population standard deviation, but do have the sample standard deviation. The Z score will be better to do as long as it is possible to do it.

No, it is called the absolute deviation.

standard deviation