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3 years ago

Exercises 1–5

Mathematics

1 answer:

Degger [83]3 years ago

6 0

Answer: The estimated value for population mean rating based on these ten sampled ratings is 7.3 .

Step-by-step explanation:

Given : A random sample of size ten produced the following ratings in the computer games rating example in the last lesson:

12, 5, 2, 4, 1, 4, 18, 10, 1, 16.

Let x denotes any value in the above data.

${\sum_{i=1}^{10}x_i}{10}=73$

We know that the sample mean $(\overline{x})$ is the best estimate to the population mean $(\mu)$ .

Sample mean= $\overline{x}=\dfrac{\sum_{i=1}^{n}x_i}{n}$

Here , n= 10

$\overline{x}=\dfrac{\sum_{i=1}^{10}x_i}{10}$

$\overline{x}=\dfrac{73}{10}=7.3$

Therefore , the estimated value for population mean rating based on these ten sampled ratings= 7.3

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

Mamont248 [21]

Answer:

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.

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3 years ago

I forgot how to do this. Please help!

Annette [7]

Answer:Here

I forgot how to do this. Please help!

Step-by-step explanation:

5 0

2 years ago

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stealth61 [152]

The answer is y=-3/2x +3 so D

8 0

3 years ago

Please help with all parts (ignore my answer for the first one)

kykrilka [37]

Please edit your question to what is the question

5 0

3 years ago

30. The perimeter of a rectangle is 292 feet. The length of the rectangle is 4 feet less than 5 times the width. What is the len

blondinia [14]

P = 2(L + W)
P = 292
L = 5W - 4

292 = 2(5W - 4 + W)
292 = 2(6W - 4)
292 = 12W - 8
292 + 8 = 12W
300 = 12W
300/12 = W
25 = W <=== width is 25 ft

L = 5W - 4
L = 5(25) - 4
L = 125 - 4
L = 121 <=== length is 121 ft

5 0

3 years ago