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3 years ago

A honeybee hive may contain up to 4.0x10^4 honeybees. What is the number of honeybees in a beehive in standard form

Mathematics

1 answer:

lilavasa [31]3 years ago

7 0

Answer:

40000

Step-by-step explanation:

Given that,

A honeybee hive may contain up to $4\times 10^4$ honeybees.

We need to find the number of honeybees in a beehive in standard form.

We know that,

10⁴ = 10000

So,

4 × 10⁴ = 4 × 10000

= 40000

So, there are 40000 honeybees in a beehive.

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Assume the sample variances to be continuous measurements. Find the probability that a random sample of 25observations, from a n

konstantin123 [22]

Answer:

a) $P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502$

b) $P(13.848$ Step-by-step explanation:

Previous concepts

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

For this case we assume that the sample variance is given by $S^2$ and we select a random sample of size n from a normal population with a population variance $\sigma^2$ . And we define the following statistic:

$T = \frac{(n-1) S^2}{\sigma^2}$

And the distribution for this statistic is $T \sim \chi^2_{n-1}$

For this case we know that n =25 and $\sigma^2 = 6$ so then our statistic would be given by:

$\chi^2 = \frac{(n-1)S^2}{\sigma^2}=\frac{24 S^2}{6}= 4S^2$

With 25-1 =24 degrees of freedom.

Solution to the problem

Part a

For this case we want this probability:

$P(S^2 > 9.1)$

And we can multiply the inequality by 4 on both sides and we got:

$P(4S^2 > 4*9.1) = P(\chi^2_{24} >36.4) = 0.0502$

And we can use the following excel code to find it: "=1-CHISQ.DIST(36.4,24,TRUE)"

Part b

For this case we want this probability:

$P(3.462 < S^2$

If we multiply the inequality by 4 on all the terms we got:

$P(3.462*4 < 4S^2 < 4*10.745)= P(13.848< \chi^2$ And we can find this probability like this: $P(13.848$ And we use the following code to find the answer in excel: "=CHISQ.DIST(42.98,24,TRUE)-CHISQ.DIST(13.848,24,TRUE)"

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3 years ago

As your sample size grows larger, the n - 1 adjustment for the standard deviation has a:______

kumpel [21]

As your sample size grows larger, the n - 1 adjustment for the standard deviation has a smaller impact on the estimates of standard deviation.

<u>Step-by-step explanation:</u>

The average (mean) of sample's distribution seems to be the same as the distribution mean at which samples were taken. The means of mean distribution will not change. However, the standard deviations for the samples mean is the standard deviations of the primary distribution divided by square roots of the samples size.

The standard deviations of means decreases as the samples size increases. Likewise, when the samples size decreases, the standard deviations for the samples mean increases. So, there is a little impact on standard deviations estimation when sample size increases.

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vovikov84 [41]

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lilavasa [31]

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Answer:

Step-by-step explanation:

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