Answer :
Answer:
[tex] P(M) = \frac{52}{390}= 0.133[/tex]
[tex] P(T) = \frac{64}{390}= 0.164[/tex]
[tex] P(W) = \frac{70}{390}= 0.179[/tex]
[tex] P(Th) = \frac{59}{390}= 0.151[/tex]
[tex] P(F) = \frac{60}{390}= 0.154[/tex]
[tex] P(S) = \frac{43}{390}= 0.110[/tex]
[tex] P(Su) = \frac{42}{390}= 0.109[/tex]
As we can see the relative frequencies are not too different the average for all the relative frequencies is [tex]\frac{1}{7}= 0.143[/tex] and the maximum absolute difference |0.143-0.109|= 0.034 is not enough higher to conclude that we have significant difference from the average or expected value.
Step-by-step explanation:
For this case we have the frequency of the number of times that the outcome happens per each day of the week:
Day Frequency
Monday 52
Tuesday 64
Wednesday 70
Thursday 59
Friday 60
Saturday 43
Sunday 42
We can begin findind the total for the frequency and we got:
52+64+70+59+60+43+42=390
And now we can calculate the relative frequency for each case like this:
[tex] P(M) = \frac{52}{390}= 0.133[/tex]
[tex] P(T) = \frac{64}{390}= 0.164[/tex]
[tex] P(W) = \frac{70}{390}= 0.179[/tex]
[tex] P(Th) = \frac{59}{390}= 0.151[/tex]
[tex] P(F) = \frac{60}{390}= 0.154[/tex]
[tex] P(S) = \frac{43}{390}= 0.110[/tex]
[tex] P(Su) = \frac{42}{390}= 0.109[/tex]
As we can see the relative frequencies are not too different the average for all the relative frequencies is [tex]\frac{1}{7}= 0.143[/tex] and the maximum absolute difference |0.143-0.109|= 0.034 is not enough higher to conclude that we have significant difference from the average or expected value.
In order to test this more formally we can use a chi square test but the problem not ask for this.