Can someone please help, ty!
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The bell curve attached below shows the normal distribution of the data.
We are looking the value of X such as the area to its left gives the probability of 0.75
We first need the z-score which we can obtain by reading from the z-table (as shown in the second picture below)
The z-score is = 0.7734
Then we use the following formula to work out X
z-score = (X - Mean) ÷ Standard Deviation
0.7734 = (X - 100) ÷ 15
0.7734×15 = X - 100
11.601 = X - 100
X = 11.601 + 100
X = 111.601 ≈ 112
Hence the third quartile is 112
We are looking the value of X such as the area to its left gives the probability of 0.75
We first need the z-score which we can obtain by reading from the z-table (as shown in the second picture below)
The z-score is = 0.7734
Then we use the following formula to work out X
z-score = (X - Mean) ÷ Standard Deviation
0.7734 = (X - 100) ÷ 15
0.7734×15 = X - 100
11.601 = X - 100
X = 11.601 + 100
X = 111.601 ≈ 112
Hence the third quartile is 112