In a sample of 11 ordered observations.the last 10 were: -0.9 0.6 0.7 1.6 2.43.1 49 54 10.6 I 0.6 The 1.5 IQR rule identified the first observation, which is missing, as an outlier. The value of the first observation could be -6.8 -6.5 -4.9

To determine if the first observation is an outlier using the 1.5 IQR rule, we need to calculate the interquartile range (IQR) and then identify the lower and upper bounds for outliers.First, let's arrange the given ordered observations in ascending order:-0.9, 0.6, 0.7, 1.6, 2.4, 3.1, 4.9, 5.4, 10.6, 11.0, 0.6Next, we need to find the median (Q2), the first quartile (Q1), and the third quartile (Q3).Since there are 11 observations, the median (Q2) is the 6th value:Q2 = 3.1To find Q1, we take the median of the first 5 values:Q1 = median of {-0.9, 0.6, 0.7, 1.6, 2.4} = 0.7To find Q3, we take the median of the last 5 values:Q3 = median of {4.9, 5.4, 10.6, 11.0, 0.6} = 5.4Now, we can calculate the IQR:IQR = Q3 - Q1 = 5.4 - 0.7 = 4.7Using the 1.5 IQR rule, the lower bound for outliers is:Lower bound = Q1 - 1.5 * IQR = 0.7 - 1.5 * 4.7 = -6.8The upper bound for outliers is:Upper bound = Q3 + 1.5 * IQR = 5.4 + 1.5 * 4.7 = 11.0Therefore, any observation below -6.8 or above 11.0 is considered an outlier.Given the options, the value of the first observation that could be an outlier is: So, the correct answer is .