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3 years ago

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A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a standard de

viation of 86 minutes with a mean life of 505 minutes. If the claim is true, in a sample of 120 batteries, what is the probability that the mean battery life would differ from the population mean by greater than 16.6 minutes? Round your answer to four decimal places.

1 answer:

Georgia [21]3 years ago

7 0

Answer:

p=0.01740

Step-by-step explanation:

-Let $\bar X$ denote the mean battery life and that it follows a normal distribution of mean 505 minutes and standard deviation of 86 minutes.

-To find the probability in a sample n=120, we calculate as follows:

$z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\=\frac{16.6}{86/\sqrt{120}}\\\\=2.1145$

The probability is thus:

$P(Z>2.1145)=1-0.9826\\\\=0.01740$

Hence, the probability is 0.01740

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Consider a set of one-dimensional points: 6, 12, 18, 24, 30, 42, 48.

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Answer:

1) With initial centroids 10-40, final clusters are

first cluster (6,12,18,24,30)

second cluster (42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 378

2) With initial centroids 10-20, final clusters are

first cluster (6,12,18,24)

second cluster (30,42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 348

Step-by-step explanation:

K-means works as follows

the points will be clustered according to their distance to the centroids.
Then centroids are updated as the cluster means.
This process continues until clusters doesn't change anymore

1) <u><em>initial centroids</em></u> 10-40

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

<u><em>new centroids</em></u> 15-40

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

<u><em>final centroids</em></u> 18-45

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

Sum of Squared errors = $(18-6)^{2}$ + $(18-12)^{2}$ + $(18-18)^{2}$ + $(18-24)^{2}$ + $(18-30)^{2}$ + $(45-42)^{2}$ + $(45-48)^{2}$ =378

2)<u><em>initial centroids</em></u> 10-20

first cluster (6,12) mean:9

second cluster (18,24,30,42,48) mean:32.4

<u><em>new centroids</em></u> 9-32.4

first cluster (6,12,18) mean:12

second cluster (24,30,42,48) mean:36

<u><em>new centroids</em></u> 12-36

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

<u><em>final centroids</em></u> 15-40

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

Sum of Squared errors = $(15-6)^{2}$ + $(15-12)^{2}$ + $(15-18)^{2}$ + $(15-24)^{2}$ + $(40-30)^{2}$ + $(40-42)^{2}$ + $(40-48)^{2}$ =348

6 0

3 years ago

The owner of two hotels is ordering towels. He bought 15 hand towels and 43 bath towels for his hotel in Livingston, spending a

coldgirl [10]

Answer:

The cost of each Hand towels is $5 and Cost of each Bath towel is $11.

Step-by-step explanation:

Let the Cost of Each hand Towels be $'x'$ .

Let the Cost of Each Bath Towels be $'y'$ .

Now Given:

15 hand towels and 43 bath towels for his hotel in Livingston, spending a total of $548.

So we can say that;

$15x+43y=548$

$15x=548-43y\\\\x=\frac{548-43y}{15}$ ⇒ Equation 1

Also Given:

99 hand towels and 62 bath towels for his hotel in Lancaster, spending $1,177.

So we can say that;

$99x+62y =1177$ ⇒ Equation 2

Substituting the value of 'x' from equation 1 in Equation 2 we get;

$99\frac{(548-43y)}{15}+62y=1177\\\\33\frac{(548-43y)}5+62y=1177\\\\\frac{18084-1419y}5+62y=1177$

Now taking LCM to make the denominator common we get;

$\frac{18084-1419y}5+\frac{62y\times5}{5}=1177\\\\\frac{18084-1419y}5+\frac{310y}5=1177\\\\\frac{18084-1419y+310y}{5}=1177$

$18084-1109y=1177\times5\\\\18084-1109y=5885$

Combining the like terms we get;

$18084-5885=1109y\\\\12199=1109y$

Dividing both side by 1109 we get;

$\frac{12199}{1109}=\frac{1109y}{1109}\\\\y=\$11$

Substituting the value of y in equation 1 we get;

$x=\frac{548-43y}{15}=\frac{548-43\times11}{15}=\$5$

Hence The cost of each Hand towels is $5 and Cost of each Bath towel is $11.

7 0

3 years ago