Only the bar for rating 4 is unknown. After you write the mean equation, solve it to find $k$.

Once you know the total number of students $N$, list the ratings in order using cumulative counts (how many 1s, then 2s, then 3s, etc.).

• If $N$ is even, the median is the average of the $\frac{N}{2}$th and $\left(\frac{N}{2}+1\right)$th values.
• If $N$ is odd, the median is the $\left(\frac{N+1}{2}\right)$th value.

Enter k=1 to create a slider.

From the bar graph, enter the mean expression as

(1*1 + 2*2 + 3*2 + 4*k + 5*1)/(1+2+2+k+1)

Graph:

• y=(1*1 + 2*2 + 3*2 + 4*k + 5*1)/(1+2+2+k+1)
• y=3.5

Move the slider for $k$ until the two horizontal lines overlap (so the mean expression equals $3.5$).

Enter N=1+2+2+k+1.

If $N$ is even, compute the middle positions:

• m1=N/2
• m2=m1+1

Then compare m1 and m2 to the cumulative counts from the graph (1s, then 2s, then 3s, then 4s) to see which rating contains both middle positions.

From the bar graph:

• Rating 1: 1 student
• Rating 2: 2 students
• Rating 3: 2 students
• Rating 4: $k$ students
• Rating 5: 1 student

The mean is

Compute the numerator and denominator:

• Numerator: $1+4+6+4k+5=16+4k$
• Denominator: $6+k$

So

Multiply both sides by $6+k$:

Subtract $3.5k$ from both sides:

So $0.5k=5$, and therefore $k=10$. (You will use this value to find the median.)

The total number of ratings is

With 16 ratings (an even number), the median is the average of the 8th and 9th values when the ratings are listed in order.

Count up from the lowest ratings:

• Ratings of 1 account for position 1
• Ratings of 2 account for positions 2 through 3
• Ratings of 3 account for positions 4 through 5

So positions 6 through 15 are ratings of 4 (because there are 10 students who rated 4).

Therefore, both the 8th and 9th values are 4, so the median is $4$.