Help me.. ASAP just number 10 pls

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Natali [406]

Answer:

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

The relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

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

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

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.

In this problem, we have that:

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.

This means that $\mu = 50$ and $\sigma = 6$ .

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

The pvalue of the zscore of $X = 60$ is the percentage of snails that take LESS than 60 hours to finish. So the percentage of snails that take more than 60 hours to finish is 100% substracted by this pvalue.

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

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

$Z = 1.67$

A Zscore of 1.67 has a pvalue of .9525. This means that there is a 95.25% of the snails take less than 60 hours to finish.

The percentage of snails that take more than 60 hours to finish is 100%-95.25% = 4.75%.

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

The relative frequence off snails that take less than 60 hours to finish is the pvalue of the zscore of X = 60.

In the item above, we find that this value is .9525.

So, the relative frequency of snails that take less than 60 hours to finish is .9525

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

This is the pvalue of the zscore of X = 67 subtracted by the pvalue of the zscore of X = 60. So

$X = 67$

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

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

$Z = 2.83$

A zscore of 2.83 has a pvalue of .9977.

For X = 60, we have found a Zscore o 1.67 with a pvalue of .9977

So, the percentage of snails that take between 60 and 67 hours to finish is:

$p = .9977 - 0.9525 = .0452$

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

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

This is 100% subtracted by the pvalue of the Zscore of X = 76.

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

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

$Z = 4.33$

The pvalue of Z = 4.33 is 1.

So, there is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

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

The most hours that a snail must finish is the value of X of the Zscore when p = 0.10.

Z = -1.29 has a pvalue of 0.0985, this is the largest pvalue below 0.1. So what is the value of X when Z = -1.29?

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

$-1.29 = \frac{X - 50}{6}$

$X - 50 = -7.74$

$X = 42.26$

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

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

This is from a pvalue of .1 to a pvalue of .9.

When the pvalue is .1, X = 42.26.

A zscore of 1.28 is the largest with a pvalue below .9. So

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

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

$X - 50 = 7.68$

$X = 57.68$

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

5 0

3 years ago

art bought 4 boxes of pens but gave his friend ben 8 pens and now has a total of 112 pens how many are in a box? which of the fo

viktelen [127]

There was 30 in each box until he gave 8 away so no there is probably 28 in each box

sorry if i am wrong

4X - 8 = 112

3 0

3 years ago

Read 2 more answers

300€ durante 5 años al 0.3% de interés compuesto mensual. A) ¿Cuánto interés se ha ganado? B) Cuánto interés se ha ganado?

marysya [2.9K]

Answer:

Interes ganado= €59.07

Step-by-step explanation:

Dada la siguiente informacion:

300€ durante 5 años al 0.3% de interés compuesto mensual.

<u>Para calcular el valor final de la inversión y el interés ganado, debemos usar la siguiente formula:</u>

FV= PV*(1 + i)^n

Present Value= 300

i= 0.003

n= 5*12= 60 meses

FV= 300*(1.003^60)

FV= 359.07

Interes ganado= 359.07 - 300

Interes ganado= 59.07

5 0

3 years ago

Explain Long division in your own words using at least 3 sentences.

Marina86 [1]

Answer:

form of devision that takes longer bc it a fraction or decimal. first you want to get rid of the decimals. then devide the numbers from the lowest to the highest

I may be wrong been for ever since I did it on paper

4 0

2 years ago

Read 2 more answers

Valerie ran Three and one-fourth laps yesterday, and her coach wants her to run Two and one-half times that many laps today. How

Maru [420]

Answer:

13/4 * 5/2

Step-by-step explanation:

In the question, it says that her coach wants her to run 2 1/2 TIMES the original 3/4 she had ran. Therefore, you would want to make the fractions into improper fractions, and then multiply, which is how you would get 13/4 * 5/2.

6 0

3 years ago