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4 years ago

Find the difference in length between a walking stick and a bumblebee

Mathematics

2 answers:

Talja [164]4 years ago

8 0

The diffrence is a walking stick is much larger then a bumblebee

hichkok12 [17]4 years ago

4 0

A walking stick is way longer than a bumble bee

You might be interested in

Find and interpret the mean absolute deviation of the data.

Zinaida [17]

Answer:

The mean absolute deviation is 4

Interpertation: The average distance between each data value and the mean is 4 which is the variation in the data set so you can understand how "spread out" each data point is from each other

Step-by-step explanation:

The mean absolute deviation is a measure of how much a data set varies.

Step 1: Compute the mean

mean = the sum of all the terms / number of terms

mean = 94 + 86 + 95 + 99 + 88 + 90 / 6

= 552 / 6

= 92

The mean is 92

Step 2: Finding the absolute deviation

You would take each number in the data set and subtract it from the mean, take the absolute value. After you take the sum of the absolute data values

= |94 - 92| + |86 - 92| + |95 - 92| + |99 - 92| + |88 - 92| + |90 - 92|

= |2| + |- 6| + |3| + |7| + |- 4| + |- 2|

= 2 + 6 + 3 + 7 + 4 + 2

= 24

To find the mean absolute deviation you divided the sum by total number of terms...

24/6 = 4

The average (mean) is equal to the sum of all the data values divided by the count of values in the data set (which was step 1)

Let me know if you have any questions !

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3 0

2 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$