The value of P(4<X≤12) is 0.45
<h3>How to determine the probability?</h3>The complete question is added as an attachment
To calculate P(4<X≤12), we use the following equation
P(4<X≤12) = P(4<X≤8) + P(8<X≤12)
From the histogram, we have:
P(4<X≤8) = 0.1
P(8<X≤12) = 0.35
So, we have:
P(4<X≤12) = 0.1 + 0.35
Evaluate
P(4<X≤12) = 0.45
Hence, the value of P(4<X≤12) is 0.45
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Answer:
And if we analyze the info provided we have 500 values
Tails: 160,165,167,171,175,...,268,269,269,269,270,270
So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.
Step-by-step explanation:
For this case we have the 5 number summary
(160,210,220,250,270)
So then we have:
minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270
If we find the interquartile range we got:
For this case we need to find the lower and upper limit with the following formulas:
And if we analyze the info provided we have 500 values
Tails: 160,165,167,171,175,...,268,269,269,269,270,270
So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.