According to a recent study, the carapace length for adult males of a certain species of tarantula are normally distributed with a mean of mu equals 18.45μ=18.45 mm and a standard deviation of sigma equals 1.63σ=1.63 mm. Complete parts (a) through (d) below.

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 %. (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 of decimal rounded to two decimal places as needed.)

Standard carapace length is a straight-line dimension and measurement which is from the anterior-most point on the centerline, The genus Grammostola are different species of big bodied tarantulas (which are usually 20 - 30 mm carapace length).

the full step by step calculation and solution is in the attached image below

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joaobezerra joaobezerra · Answer 2 · 2021-12-03T14:51:15+01:00

Using the normal distribution, it is found that:

a) The percentage of the tarantulas that have a carapace length between 15 and 16 is 4.98%.

b) The percentage of the tarantulas that have a carapace length exceeding 18 is 60.87%.

c)

The first quartile is of 17.33 mm.

The third quartile is of 19.53 mm.

The interpretation is that the middle 50% of the measures is between 17.33 mm and 19.53 mm.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is [tex]\mu = 18.45[/tex]
The standard deviation is [tex]\sigma = 1.63[/tex]

Item a:

The proportion is the p-value of Z when X = 16 subtracted by the p-value of Z when X = 15.

X = 16:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{16 - 18.45}{1.63}[/tex]

[tex]Z = -1.5[/tex]

[tex]Z = -1.5[/tex] has a p-value of 0.0668

X = 15:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{15 - 18.45}{1.63}[/tex]

[tex]Z = -2.12[/tex]

[tex]Z = -2.12[/tex] has a p-value of 0.017

0.0668 - 0.017 = 0.0498

0.0498 x 100% = 4.98%

The percentage of the tarantulas that have a carapace length between 15 and 16 is 4.98%.

Item b:

The proportion is 1 subtracted by the p-value of Z when X = 18, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{18 - 18.45}{1.63}[/tex]

[tex]Z = -0.276[/tex]

[tex]Z = -0.276[/tex] has a p-value of 0.3913.

1 - 0.3913 = 0.6087

0.6087 x 100% = 60.87%

The percentage of the tarantulas that have a carapace length exceeding 18 is 60.87%.

Item c:

The first quartile is the 25th percentile, which is X when Z has a p-value of 0.25, so X when Z = -0.675.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.675 = \frac{X - 18.45}{1.63}[/tex]

[tex]X - 18.45 = -0.675(1.63)[/tex]

[tex]X = 17.33[/tex]

The first quartile is of 17.33 mm.

The third quartile is the 75th percentile, which is X when Z has a p-value of 0.25, so X when Z = 0.675.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.675 = \frac{X - 18.45}{1.63}[/tex]

[tex]X - 18.45 = 0.675(1.63)[/tex]

[tex]X = 19.53[/tex]

The third quartile is of 19.53 mm.

The interpretation is that the middle 50% of the measures is between 17.33 mm and 19.53 mm.

A similar problem is given at https://brainly.com/question/24663213