7 2/15 + 5 2/3 + 9 13/15
1 answer:
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Answer:
n = 66.564
Step-by-step explanation:
- Because the population is unknown, we will apply the following formula to find the sample size:
Where:
z = confidence level score.
S = standard deviation.
E = error range.
2. We will find each of these three data and replace them in the formula.
"z" theoretically is a value that measures how many standard deviations an element has to the mean. For each confidence level there is an associated z value. In the question, this level is 99%, which is equivalent to a z value of 2.58. To find this figure it is not necessary to follow any mathematical procedure, it is enough to make use of a z-score table, which shows the values for any confidence interval.
The standard deviation is already provided by the question, it is S = 100.
Finally, "E" is the acceptable limit of sampling error. In the example, we can find this data. Let us note that in the end it says that the director wishes to estimate the mean number of admissions to within 1 admission, this means that she is willing to tolerate a miscalculation of just 1 admission.
Once this data is identified, we replace in the formula:
3. The corresponding mathematical operations are developed:
n= 66.564
Answer:
Step-by-step explanation:
Since, The total number of student = 300
Out of which,
The number of students who are only in Maths = 120
And, The number of students who are only in Science = 50
While, the students who are not from any subject = 100
Hence, the number of student who are from both maths and science = Total student - Maths student (only) - science student (only) - None
= 300 - 120 - 50 - 100
= 30
That is, there are 30 students who are both from science and maths,
Thus, the probability of selecting one student who is both from maths and science = 30/300 = 1/10