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Q: N equals 36 with a population mean of 74 and a mean score of 79.4 with a standard deviation of 18?

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Bob scored 300 or 700.

The absolute value of the standard score becomes smaller.

A z-score requires the mean and standard deviation (or standard error). There is, therefore, not enough information to answer the question.

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

If the Z-Score corresponds to the standard deviation, then the distribution is "normal", or Gaussian.

Related questions

Bob scored 300 or 700.

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

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.

The z score, for a value y, is (y - 18.6)/4

1.41

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

The standardised score decreases.

score of 92

The absolute value of the standard score becomes smaller.

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.

standard deviation

z-score or standard score... tells you how many standard deviations away from the mean a particular number is in relations to all numbers in a population (or sample)

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