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Calculus and Above

I need help. I'm trying to determine the likelihood of a

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I need help. I'm...
I need help. I'm trying to determine the likelihood of a response in thresholds (ranges) given historical data. For example: On a survey that has a 100 data points, with a mean score of 60 and a standard deviation of 2 what are probabilities of us achieving a 70? Or 65? or 68? (Within what time frame?)
Submitted: 1 year ago.Category: Calculus and Above
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3/30/2016
Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago
Rajan.Sareen, Bachelor's Degree
Category: Calculus and Above
Satisfied Customers: 86
Experience: I am a expert with more than 4 years of expreience.
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Hi, Welcome to just answer. I will be answering your question.

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Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago

These types of problem can be solved using normal distribution.
here we have

Mean = 60 and SD = 2, n = 100

Z score is given as (X-Mean)/(SD/sqrt(n))

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Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago

We need to find

P(X=70)

Since here mean and standard deviation are based on 100 data points so standard deviation will not be divided by sqrt(n)

P((X-mean)/(SD)=(70-60)/(2)

P(Z=10/(2)

P(Z=5) = 0.000

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Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago

P(X=65)

Since here mean and standard deviation are based on 100 data points so standard deviation will not be divided by sqrt(n)

P((X-mean)/(SD)=(65-60)/(2)

P(Z=5/(2)

P(Z=2.5) = 0.99379

0.99379 value can be found from standard normal table with Z value equals to 2.5.

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Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago

P(X=68)

Since here mean and standard deviation are based on 100 data points so standard deviation will not be divided by sqrt(n)

P((X-mean)/(SD)=(68-60)/(2)

P(Z=8/(2)

P(Z=4) =0.999968

0.999968 value can be found from standard normal table with Z value equals to 4.

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Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago

Ignore the above part for X= 70, use the below as I pasted 0.000 instead of 1.000.

P(X=70)

Since here mean and standard deviation are based on 100 data points so standard deviation will not be divided by sqrt(n)

P((X-mean)/(SD)=(70-60)/(2)

P(Z=10/(2)

P(Z=5) = 1.000

1.000 value can be found from standard normal table with Z value equals to 5.

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Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago

Please let me know if you have any query, I will be happy to help you. Thanks!

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Customer reply replied 1 year ago

Rajan,

I am no mathematician can you explain in lehman's terms and tell me what Z value is? I'm assuming P = Probability and X = whatever number I'd trying to determine the probability of hitting that # ***** this scale?

Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago

Z is the standardized value of normal distribution. Z can be calculated as (X-Mean)/(SD/sqrt(n)) if we know the mean, standard deviation and n, we can find Z value so that we can calculate the probability. We have standardized value for Z known as standard normal probability tables. The table can be found from below link for corresponding value of Z:

http://www.stat.purdue.edu/~mccabe/ips4tab/bmtables.pdf

P is probability and x is the number we are interested in finding.

Please let me know if you have any query. I will be happy to help you. Thanks!

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Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago

Hi, You are requested to rate my service and answer. Thanks!

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Tutor: Rajan.Sareen, Bachelor's Degree replied 1 year ago

Thanks!

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