B. 21.8°

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

Sunshine High School claims that 72% of its students complete their homework on a daily basis. Ms. Carter surveyed 80 students a

Ulleksa [173]

Answer: z equals 0.60 minus 0.72 all over the square root of the quantity 0.72 times 0.28 over 80

Step-by-step explanation:

let p be the population proportion of students complete their homework on a daily basis.

As per given , we have

p= 0.72

sample size : n= 80

Sample proportion : $\hat{p}=0.60$

Test statistic for proportion:

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

i.e.

$z=\dfrac{0.60-0.72}{\sqrt{\dfrac{0.72(1-0.72)}{80}}}$

i.e.

$z=\dfrac{0.60-0.72}{\sqrt{\dfrac{0.72(0.28)}{80}}}$

That is , z equals 0.60 minus 0.72 all over the square root of the quantity 0.72 times 0.28 over 80

6 0

3 years ago

Is (3,-2) a solution to the system below? How do you know?

cricket20 [7]

Step-by-step explanation:

you know by using the x and y coordinates of the point in the equations and see, if both equations are still true.

if yes, it was a solution, and if just only one is false, then it was not a solution.

(3, -2) in

y + 4x = 10

5x - y = 13

so,

-2 + 4×3 = 10

-2 + 12 = 10

10 = 10 true

5×3 - (-2) = 13

15 + 2 = 13

17 = 13 false

so, it was NOT a solution.

3 0

2 years ago

70/9 as a mixed number

Makovka662 [10]

70/9
Divide
9*7 = 63
70-63 = 7

7 7/9

7 0

3 years ago

Read 2 more answers

Ian's monthly allowance is $21. In January he starts saving for a birthday gift in June. Each month he saves of his allowance. T

Dmitrij [34]

Answer:

<h2>Ian will have enough money to buy the gift</h2>

Step-by-step explanation:

Step one:

given

monthly allowance= $21

from January to June= 6 months

total savings = 21*6= $126

cost of gift = $110

Step two:

Ian will have enough money to buy the gift

since he saved $126 which is greater than the cost of the gift of $110

he would have a balance of $16

4 0

3 years ago