According to a recent study, the carapace length for adult males of a certain sp

Question

According to a recent study, the carapace length for adult males of a certain species of tarantula are normally distributed with a mean of = 18.86 mm and a standard deviation of = 1.74 mm. Complete parts (a) through (d) below lick here to vie the standard normal distribution table the standard normal distribution table ick here to VI a. Find the percentage of the tarantulas that have a carapace length between 17 mm and 18 mnm The percentage of the tarantulas that have a carapace length between 17 and 18 is % (Type an integer or decimal rounded to two decimal places as needed.) b. Find the percentage of the tarantulas that have a carapace length exceeding 20 mm The percentage of the tarantulas that have a carapace length exceeding 20 is % (Type an integer or decimal rounded to two decimal places as needed.) c. Determine and interpret the quartiles for the carapace length of these tarantulas The first quartile is (Type an integer or decimal rounded to two decimal places as needed.) Interpret the first quartile. Select the correct choice below and fill in the answer box(es) to complete your choice Type integers or decimals rounded to two decimal places as needed.) 0 A. This first quartile is % less than the average carapace length % greater than the smallest carapace length B. This first quartile is This first quartile means that % of the carapace lengths are equal to 0 D. This first quartile means that % of the carapace lengths are less than The second quartile is (Type an integer or decimal rounded to two decimal places as needed.)

Explanation / Answer

a)

b)

=25.62%

c) for first quartile z score =-0. 6745

hence corresponding lenght =mean+z*std deviation =20.03

option D is correct : the first quartile means that 25% of the carapace length are less than 20.03

second quartile is 18.86

for normal distribution z score =(X-)/

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