Answer:
a) Lower inner fence = 77.6168 = 77.62 to 2 d.p
Lower outer fence = 48.9044 = 48.90 to 2 d.p
b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74) is 0.00159
Step-by-step explanation:
Lower inner and outer fences are used to illustrate or write off extreme values of a data set (the outliers).
Lower inner fence = Q₁ – (1.5 × IQR)
Lower outer fence = Q₁ – (3 × IQR)
Q₁ = 25th percentile = lower quartile
IQR = Inter quartile Range = Q₃ - Q₁
Q₃ = 75th percentile = upper quartile
To calculate Q₁ for a normal distribution with only mean and standard deviation known,
We need the standardized score whose probability is 0.25 P(z) = 0.25
From the normal distribution table
z = (± 0.674)
z = (x - xbar)/σ
x = the value in the data we're interested in,
xbar = mean = 115.9
σ = standard deviation = 14.2
Lower quartile corresponds to (z = - 0.674)
- 0.674 = (x - 115.9)/14.2
Q₁ = X = 106.3292
The upper quartile, Q₃ corresponds to z = (+0.674)
Q₃ = 125.4708
IQR = 125.4708 - 106.3292 = 19.1416
Lower inner fence = Q₁ – (1.5 × IQR)
Lower outer fence = Q₁ – (3 × IQR)
Lower inner fence = 106.3292 - (1.5 × 19.1416) = 106.3292 - 28.7124 = 77.6168
Lower outer fence = 106.3292 – (3 × 19.1416) = 48.9044
b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74)
We standardize 74 by obtaining its z-score
z = (x - xbar)/σ
z = (74 - 115.9)/14.2 = - 2.95
P(x ≤ 74) = P(z ≤ -2.95) = 0.00159 (Obtained from normal distribution tables)
