Respuesta :
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
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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