Mathematics, 14.03.2020 04:24 tylerwoodson

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean 50 hours and standard deviation 6 hours. For the following questions, use the normal chart and the method in class. The correct procedure results in precisely the required accuracy. a. The percentage of snails that take more than 60 hours to finish is %.b. The relative frequency of snails that take less than 60 hours to finish is c. The proportion of snails that take between 60 and 67 hours to finish is d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places) is e. To be among the 10% fastest snails, a snail must finish in at most hours. f. The most typical 80% of snails take between and hours to finish.

Answers: 2

Show answers

Answers

Answer from: dieguezisabel

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

Z = 1.67

Z = 1.67 has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

Z = 1.67

Z = 1.67 has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

Z = 2.83

Z = 2.83 has a pvalue 0.9977

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

Z = 1.67

Z = 1.67 has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

Z = 4.33

Z = 4.33 has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

X - 50 = -1.28*6

X = 42.32

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

X - 50 = -1.28*6

X = 42.32

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

X - 50 = 1.28*6

X = 57.68

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Answer from: Quest

last one.

step-by-step explanation:

last one.

generally speaking, the others follow some sort of pattern that correlates to either linear (looks like a straight line) or sinusoidal (looks like an s).

the last one is just a random cloud of points that doesn't correlate to any particular known order.

Answer from: Quest

step-by-step explanation:

10 ptsfind the area of the inside of the frame where the picture would belong. then find the area of

Another question on Mathematics

Mathematics, 21.06.2019 15:00

12 point are triangles pqr and stu congruent? what is the congruency that proves they are congruent? what is the perimeter of triangle pqr?

Answers: 2

Answer

Mathematics, 21.06.2019 20:00

Karen will divide her garden into equal parts. she will plant corn in 8/12 of the garden. what is the lowest number of parts she can divide her garden into?

Answers: 1

Answer

Mathematics, 22.06.2019 00:40

Afootball quarterback has two more chances to throw a touchdown before his team is forced to punt the ball. he misses the receiver on the first throw 25% of the time. when his first throw is incomplete, he misses the receiver on the second throw 15% of the time. what is the probability of not throwing the ball to a receiver on either throw?

Answers: 2

Answer

Mathematics, 22.06.2019 02:00

Elm street is straight. willard's house is at point h between the school at point s and the mall at point m. if sh = 3 miles and hm = 4.5 miles, what is the value of sm in miles?

Answers: 1

Answer

You know the right answer?

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately no...

Questions