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astra-53 [7]
3 years ago
7

Braden wants to compare the types of ducks in two nearby ponds. He collected two samples using the following criteria. For sampl

e 1, he counted the types of ducks on Walden Pond from 6 A.M. to 9 A.M. For sample 2, he counted the number of Mallard ducks that were on Brook’s Pond from 12 P.M. to 3 P.M. What would improve Braden’s plan? Check all that apply.
Mathematics
2 answers:
Zinaida [17]3 years ago
5 0

Answer:

1 and 4

Step-by-step explanation:


Sladkaya [172]3 years ago
4 0

Answer:

Step-by-step explanation:

Braden wants to compare the types of ducks in two nearby ponds. He collected two samples using the following criteria. For sample 1, he counted the types of ducks on Walden Pond from 6 A.M. to 9 A.M. For sample 2, he counted the number of Mallard ducks that were on Brook’s Pond from 12 P.M. to 3 P.M. What would improve Braden’s plan? Check all that apply.

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The mean life span of a brand name tire is 50,000 miles. Assume that the life spans of the tires are normally distributed, and t
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Answer:

a) P(X

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the mean life span of a brand name tire, and for this case we know the distribution for X is given by:

X \sim N(\mu=50000,\sigma=800)  

Part a

We want this probability:

P(X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X

Part b

Let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(50000,\frac{800}{\sqrt{100}})

We want this probability:

P(\bar X>50200)=1-P(\bar X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

If we apply this formula to our probability we got this:

P(\bar X >50200)=1-P(Z

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