For the purposes of constructing modified​ boxplots, outliers are defined as data values that are above Q3 by an amount greater than 1.5×IQR or below Q1 by an amount greater than 1.5×IQR​, where IQR is the interquartile range. Using this definition of​ outliers, find the probability that when a value is randomly selected from a normal​ distribution, it is an outlier.

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Answer:

0.0602

Step-by-step explanation:

Given that a value is randomly selected from a normal​ distribution at random.

To find the probability that the value is an outlier

Z score (std) for Q1 = 0.675 and Q3 ==0.675

IQR = 1.250

1.5xIQR =1.875

P(outlier) = P(Z<-1.875 or Z>1.875)

=2(0.5-0.4699)

= 0.0602

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Values which lies below the point Q1 - (1.5 × IQR) and above the point Q3 + (1.5 × IQR) are called outliers, the probability of randomly selecting an outlier value from a normal distribution is 0.0069757

Interquartile Range (IQR) = Q3 - Q1

Q3 = 75th Percentile

Q1 = 25th Percentile

Using a normal distribution table :

  • Q3 ; Zscore of 0.75 = 0.675
  • Q1 ; Zscore of 0.25 = - 0.674

Zscore of IQR = (0.675 - (-0.674)) = 0.1349

Upper limit = Q3 + (1.5 × IQR) = 0.675 + (1.5 × 0.1349) = 0.675 + 2.0235 = 2.6985

Lower limit = Q1 - (1.5 × IQR) = - 0.674 - (1.5 × 0.1349) = - 0.674 - 2.0235 = - 2.6975

Calculating the probability for each limit :

Lower limit : P(Z < - 2.6975) = 0.0034931

Upper limit : P(Z > 2.6985) = 1 - P(Z < 2.6985) = 1 - 0.99652 = 0.0034826

P(outlier) = Lower limit + Upper limit

P(outlier) = P(Z < - 2.6975) + P(Z > 2.6985)

P(outlier) = 0.0034931 + 0.0034826 = 0.0069757

Therefore, the probability of randomly selecting an outlier from a normal distribution is 0.0069757

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