Question

at a large university, 20% of students are enrolled in the nursing program. the dean of students selects a random sample of 20 students and records ( n ) = the number of students enrolled in the nursing program. the dean decides to simulate this random process by using a random number table. he assigns the digits to the outcomes.
1, 2 = student enrolled in nursing program
3 - 9, 0 = student not enrolled in nursing program
here is a portion of a random number table.
table of random digits
1 31645 03495 96193 10898 88532 73869
2 67940 85019 98036 98252 43838 45644
3 21805 26727 73239 53929 42564 17080
in one trial of this simulation, 3 of the 20 randomly selected students are enrolled in the nursing program. use line 2 to complete a second trial and then use line 3 to complete a third trial. based on the results of these 3 simulations, what is the average number of students expected to be enrolled in the nursing program in random samples of size 20?

Explanation:

Step1: Count students in line 2

Count digits \(1\) or \(2\) in line \(2\): \(67940\ 85019\ 98036\ 98252\ 43838\ 45644\).
Digits: \(6,7,9,4,0,8,5,0,1,9,9,8,0,3,6,9,8,2,5,2\).
Number of \(1\)s and \(2\)s: \(1,2,2\). So \(n_2 = 3\).

Step2: Count students in line 3

Count digits \(1\) or \(2\) in line \(3\): \(21805\ 26727\ 73239\ 53929\ 42564\ 17080\).
Digits: \(2,1,8,0,5,2,6,7,2,7,7,3,2,3,9,5,3,9,2,9,4,2,5,6,4,1,7,0,8,0\).
Number of \(1\)s and \(2\)s: \(2,1,2,2,2,2,2,1\). So \(n_3 = 8\).

Step3: Calculate average

We know \(n_1 = 3\) (given), \(n_2 = 3\), \(n_3 = 8\).
Average \(\bar{n}=\frac{n_1 + n_2 + n_3}{3}=\frac{3 + 3+8}{3}=\frac{14}{3}\approx4.67\). But wait, let's re - check line 2 and line 3 properly.

Re - check line 2:
Line 2: \(67940\ 85019\ 98036\ 98252\ 43838\ 45644\)
Digits:

• First group \(6,7,9,4,0\): no \(1\) or \(2\)
• Second group \(8,5,0,1,9\): one \(1\)
• Third group \(9,8,0,3,6\): no \(1\) or \(2\)
• Fourth group \(9,8,2,5,2\): two \(2\)s
• Fifth group \(4,3,8,3,8\): no \(1\) or \(2\)
• Sixth group \(4,5,6,4,4\): no \(1\) or \(2\)

\(n_2=1 + 2=3\)

Re - check line 3:
Line 3: \(21805\ 26727\ 73239\ 53929\ 42564\ 17080\)
Digits:

• First group \(2,1,8,0,5\): one \(2\), one \(1\)
• Second group \(2,6,7,2,7\): two \(2\)s
• Third group \(7,3,2,3,9\): one \(2\)
• Fourth group \(5,3,9,2,9\): one \(2\)
• Fifth group \(4,2,5,6,4\): one \(2\)
• Sixth group \(1,7,0,8,0\): one \(1\)

\(n_3=(1 + 1)+2 + 1+1+1+1=8\)

The three values are \(n_1 = 3\), \(n_2 = 3\), \(n_3 = 4\) (wait, no, re - check line 3 again. Wait, the problem says "use line 2 to complete a second trial and then use line 3 to complete a third trial". Maybe there was a mis - count before.

Let's count line 2 again:
Line 2: \(67940\ 85019\ 98036\ 98252\ 43838\ 45644\)

• \(6,7,9,4,0\): 0
• \(8,5,0,1,9\): 1
• \(9,8,0,3,6\): 0
• \(9,8,2,5,2\): 2
• \(4,3,8,3,8\): 0
• \(4,5,6,4,4\): 0. So \(n_2=1 + 2=3\)

Line 3: \(21805\ 26727\ 73239\ 53929\ 42564\ 17080\)

• \(2,1,8,0,5\): 2 (digits \(2\) and \(1\))
• \(2,6,7,2,7\): 2 (two \(2\)s)
• \(7,3,2,3,9\): 1 (\(2\))
• \(5,3,9,2,9\): 1 (\(2\))
• \(4,2,5,6,4\): 1 (\(2\))
• \(1,7,0,8,0\): 1 (\(1\))

\(n_3=2 + 2+1+1+1+1 = 8\). No, wait the problem may have a typo. Wait, if we consider each digit in the 20 - digit sample (since sample size \(n = 20\)).

Line 2: we take 20 digits. Let's write out all digits in line 2: \(6,7,9,4,0,8,5,0,1,9,9,8,0,3,6,9,8,2,5,2\). The number of \(1\)s and \(2\)s: \(1,2,2\) → \(n_2 = 3\)

Line 3: digits \(2,1,8,0,5,2,6,7,2,7,7,3,2,3,9,5,3,9,2,9,4,2,5,6,4,1,7,0,8,0\). Wait no, we need 20 digits. Let's take first 20 digits of line 3: \(2,1,8,0,5,2,6,7,2,7,7,3,2,3,9,5,3,9,2,9\). The number of \(1\)s and \(2\)s: \(2,1,2,2,2,2\) → \(n_3 = 6\)

Now, with \(n_1 = 3\) (given), \(n_2 = 3\), \(n_3 = 6\)

Average \(\bar{n}=\frac{3 + 3+6}{3}=\frac{12}{3}=4\)

Answer:

\(4\)