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egoroff_w [7]
3 years ago
12

A frog is sitting exactly in the middle of a board that is five feet long. Every ten seconds he jumps one foot to the left or on

e foot to the right, at random. You want to use simulation to estimate the probability that he jumps off the board in sixty seconds or less. (For example, if he jumps LLRLRL, he is still on the board, on the left-most square. If he jumps LLRLL, he has jumped off the board in fifty seconds.)
Required:
Describe how you would use the digits of a random number table to carry out this simulation.
Mathematics
1 answer:
mart [117]3 years ago
4 0

Answer:

Since the Frog makes a Jump every 10 seconds

   ( using the digits on a random number table )

  • It will take 7 jumps with 2 jumps in reverse direction i.e. either left or right for the frog to be off the board in 60 seconds
  • it will take 3 jumps in similar direction as well to get the frog off the board
  • also 5 jumps with 1 jump in the opposite direction is what it will take to get the frog off the board as well within the time frame of 60 seconds

Step-by-step explanation:

A frog sitting exactly in the middle of a board which is 5ft long : frog is 2.5 ft from the edge of the board

every 10 seconds the frog jumps ; L or R

If he jumps ; LLRLRL ( still on Board on the left-most square )

if he jumps ; LLRLL ( He is off the Board in fifty seconds )

Since the Frog makes a Jump every 10 seconds

   ( using the digits on a random number table )

  • It will take 7 jumps with 2 jumps in reverse direction i.e. either left or right for the frog to be off the board in 60 seconds
  • it will take 3 jumps in similar direction as well to get the frog off the board
  • also 5 jumps with 1 jump in the opposite direction is what it will take to get the frog off the board as well within the time frame of 60 seconds
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Step-by-step explanation:

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3 years ago
An SRS of 25 recent birth records at the local hospital was selected. In the sample, the average birth weight was x = 119.6 ounc
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Answer:

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

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

\mu_{\bar X}= \mu = 119.6

And now for the deviation we have this:

SE_{\bar X} = \frac{6.5}{\sqrt{25}}=1.3

So then the correct answer for this caee would be:

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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 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".

Solution to the problem

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

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

\mu_{\bar X}= \mu = 119.6

And now for the deviation we have this:

SE_{\bar X} = \frac{6.5}{\sqrt{25}}=1.3

So then the correct answer for this caee would be:

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2 years ago
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julia-pushkina [17]

Answer:

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Step-by-step explanation:

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7 0
2 years ago
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