Question

A die is thrown 120 times and the
frequencies of various faces are as
follow: Face No 1 2 3 4 5 6
Frequency 10 15 25 25 18 27 Find the statistical value​

Answer 1

Answer:

Suppose we wish to determine if an ordinary-looking six-sided die is fair, or balanced, meaning that every face has probability 1/6 of landing on top when the die is tossed. We could toss the die dozens, maybe hundreds, of times and compare the actual number of times each face landed on top to the expected number, which would be 1/6 of the total number of tosses. We wouldn’t expect each number to be exactly 1/6 of the total, but it should be close. To be specific, suppose the die is tossed n = 60 times with the results summarized in Table 11.8 "Die Contingency Table". For ease of reference we add a column of expected frequencies, which in this simple example is simply a column of 10s. The result is shown as Table 11.9 "Updated Die Contingency Table". In analogy with the previous section we call this an “updated” table. A measure of how much the data deviate from what we would expect to see if the die really were fair is the sum of the squares of the differences between the observed frequency O and the expected frequency E in each row, or, standardizing by dividing each square by the expected number, the sum

Answer 2

Answer:  The expected statistical value​ is 3.89

Step-by-step explanation:

P(1) ⇒ Probability of getting 1 on dice  $= \frac{10}{120}$

P(2) ⇒ Probability of getting 2 on dice  $= \frac{15}{120}$

P(3) ⇒ Probability of getting 3 on dice  $= \frac{25}{120}$

P(4) ⇒ Probability of getting 4 on dice  $= \frac{25}{120}$

P(5) ⇒ Probability of getting 5 on dice  $= \frac{18}{120}$

P(6) ⇒ Probability of getting 6 on dice  $= \frac{27}{120}$

Expected Statistical Value = Σ x.P(x)

Expected Statistical Value = 1.P(1) + 2.P(2) + 3.P(3) + 4.P(4) + 5.P(5) + 6.P(6)

Expected Statistical Value =  1/12 + 1/4 + 5/8 + 5/6+  3/4 + 27/20

Expected Statistical Value =  467/120 = 3.89

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