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 = 17.39 and a standard deviation of o 1.78 mm. Complete parts (a) through (d) below Click here to view page 1 of the standard normal distribution table a. Find the percentage of the tarantulas that have a carapace length between 15 mm and 16 mm The percentage of the tarantulas that have a carapace length between 15 and 16 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 18 mm. The percentage of the tarantulas that have a carapace length exceeding 18 is 96. 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.) O A. This first quartile means that% of the carapace lengths are equal to B. Thiq firefnilartile means that 1%, nf thecaranace len ths are less than

Explanation / Answer

a)

b)

c)for 25th percentile z score =-0.67

first quartile =mean +z*Std deviaiton =16.20

( please try 16.19 if above does not work due to rounding)

option B ) first quartile means that 25% of the carapace lengths are less than 16.20

for normal distribution z score =(X-)/ here mean=       = 17.39 std deviation   == 1.780

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