Step 3: Place parentheses around the numbers above and below the median. Not necessary statistically, but it makes Q1 and Q3 easier to spot. (1, 2, 5, 6, 7), 9, (12, 15, 18, 19, 27).
Step 4: Find Q1 and Q3 Think of Q1 as a median in the lower half of the data and think of Q3 as a median for the upper half of data. (1, 2, 5, 6, 7), 9, ( 12, 15, 18, 19, 27). Q1 = 5 and Q3 = 18.
Step 5: Subtract Q1 from Q3 to find the interquartile range. 18 – 5 = 13.
We have been given that final exam scores are normally distributed with a mean of 74 and a standard deviation of 6.
First of all we will find z-score using z-score formula.
Now let us find z-score for 86.
Now we will find P(-1<Z) which is probability that a random score would be greater than 68. We will find P(2>Z) which is probability that a random score would be less than 86.
Using normal distribution table we will get,
We will use formula to find the probability to find that a normal variable lies between two values.
Upon substituting our given values in above formula we will get,
Upon converting 0.81859 to percentage we will get
Therefore, 81.86% of final exam score will be between 68 and 86.