How to work out expected frequency from normal distribution

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SUMMARY

The expected frequency for intervals of trains can be calculated using the normal distribution by adjusting the interval boundaries to account for continuity. For example, the interval {60 ≤ trains ≤ 62} should be treated as {59.5 ≤ trains ≤ 62.5}. Using this adjustment, the expected frequency can be derived from the normal distribution table. In this case, the expected frequency for the interval (59.5, 62.5) is approximately 4.178, which aligns closely with the tabulated value of 4.13. Accurate computations can be performed using tools like Maple, which provide precise results compared to cruder methods.

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  • Understanding of normal distribution and Z-scores
  • Familiarity with expected frequency calculations
  • Basic knowledge of statistical tools like Maple
  • Concept of continuity correction in statistical distributions
NEXT STEPS
  • Learn how to apply continuity correction in normal distribution calculations
  • Explore the use of Maple for statistical computations
  • Study goodness-of-fit tests and their application in statistics
  • Investigate the differences between discrete and continuous distributions
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Statisticians, data analysts, and anyone involved in statistical modeling or frequency analysis will benefit from this discussion.

question dude
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How is the expected frequency column worked out for each interval of trains?

2) My attempt

Take the first interval, 60 - 62, I thought about doing this:

(62 - mean) / standard deviation

(62 - 67.45) / 2.92 = - 1.866

using Z score < - 1.886, from the normal distribution table, I get:

1 - 0.9686 = 0.0314

0.0314*(100) = 3.14

please note 100 is the total observed frequency

As you can see, I get 3.14 instead of 4.13 as given in the expected frequency column.
 
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question dude said:
attachment.php?attachmentid=455065&d=1440381068.jpg
How is the expected frequency column worked out for each interval of trains?

2) My attempt

Take the first interval, 60 - 62, I thought about doing this:

(62 - mean) / standard deviation

(62 - 67.45) / 2.92 = - 1.866

using Z score < - 1.886, from the normal distribution table, I get:

1 - 0.9686 = 0.0314

0.0314*(100) = 3.14

please note 100 is the total observed frequency

As you can see, I get 3.14 instead of 4.13 as given in the expected frequency column.

I don't get your answers; I don't get the tabulated expected frequencies, either, but I come close to the latter.

The number of trains is integer-valued (i.e, whole numbers) but you are approximating its distribution by a continuous distribution (the normal). So, the statement {60 ≤ trains ≤ 62} is the same as {59.5 ≤ trains ≤ 62.5} for actual, physical trains. If you use the normal distribution on the interval (59.5,62.5) you will get an expected frequency of 100* 0.04178 ≈ 4.178, which is not that far from the tabulate value of 4.13. For the interval (63 → 65 ) = (62.5 → 65.5) I get an expected frequency of 100 * 0.2071 = 20.71, which is close to the tabulated 20.68.

I used Maple to do accurate computations; if the tabulator used cruder tools he/she could get less accurate answers.

BTW: in goodness-of-fit tests we do NOT usually round off the "expected frequencies" to whole numbers; for the first cell we would typically leave it as 4.178 (or maybe 4.18, or maybe 4.2). The reason for this is that there is no reason at all to assume the expected frequencies to be integers. This has nothing at all to do with whether or not the distribution is discrete (for whole numbers) or for a continuous (like the normal): the expected cell frequency for an integer-valued random variable can --- and usually is --- a non-integer quantity.
 

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