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2 answers:
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Answer:
0.0104
Step-by-step explanation:
Given that, the lifespans of crocodiles have an approximately Normal distribution, with a mean of 18 years and a standard deviation of 2.6 years.
So, mean,
and standard deviation,
The z-score, for any arbitrary life span of x year, is given by
By using the given values, we have
The z-score for x=24, by using equation (i), is
From the table, the area to the left side of z=2.31 =0.98956
But, the proportion of crocodiles have lifespans of at least 24 years
= Area to the right side of the z=2.31
=1- (Area to the left side of the z=2.31)
=1-0.98956
=0.01044
So, the proportion of crocodiles that have lifespans of at least 24 years is 0.0104.
Hence, option (a) is correct.
The IQR is 42.5
Step-by-step explanation:
Interquartile range is the difference of third and first quartile.
First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.
As the number of values are odd
n=21
The median will be:
Putting n=21
The 11th term is 133
So median = 133
Now the data is divided into two halves
One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132
2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180
Q1 will be the median of first half and Q3 will be the median of 2nd half.
As now the halves contain even number of values, the medians will be the average of middle two values
<u>For First Half:</u>
98, 100, 101, 102, <u>108, 109</u>, 111,118, 129, 132
<u>For Second Half:</u>
135, 135, 145, 146, 146, 156, 170 176, 180, 180
Now
<u>Interquartile Range:</u>
Hence,
The IQR is 42.5
Keywords: Median, IQR
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