Question

Suppose x has a distribution with a mean of 80 and a standard deviation of 52. Random samples of size n = 64 are drawn. (a) Describe the x distribution and compute the mean and standard deviation of the distribution. x has an approximately normal distribution with mean μx = 80 and standard deviation σx = 52. (b) Find the z value corresponding to x = 86.5. z = (c) Find P(x < 86.5). (Round your answer to four decimal places.) P(x < 86.5) = (d) Would it be unusual for a random sample of size 64 from the x distribution to have a sample mean less than 86.5? Explain. Yes, it would be unusual because less than 5% of all such samples have means less than 86.5. No, it would not be unusual because less than 5% of all such samples have means less than 86.5. No, it would not be unusual because more than 5% of all such samples have means less than 86.5. Yes, it would be unusual because more than 5% of all such samples have means less than 86.5.

          Suppose x has a distribution with a mean of 80 and a standard deviation of 52. Random samples of size n = 64 are drawn.

(a) Describe the x distribution and compute the mean and standard deviation of the distribution. 
x has an approximately normal distribution with mean μx = 80 and standard deviation σx = 52.

(b) Find the z value corresponding to x = 86.5.
z = 

(c) Find P(x < 86.5). (Round your answer to four decimal places.)
P(x < 86.5) = 

(d) Would it be unusual for a random sample of size 64 from the x distribution to have a sample mean less than 86.5? Explain.
Yes, it would be unusual because less than 5% of all such samples have means less than 86.5.
No, it would not be unusual because less than 5% of all such samples have means less than 86.5.
No, it would not be unusual because more than 5% of all such samples have means less than 86.5.
Yes, it would be unusual because more than 5% of all such samples have means less than 86.5.

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Added by Brittany Y.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Ahmet Yavas

11/06/2022

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5 (calculated as 52 / √64 = 6.5). ** Show more…

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Suppose x has a distribution with a mean of 80 and a standard deviation of 52. Random samples of size n = 64 are drawn. (a) Describe the x distribution and compute the mean and standard deviation of the distribution. x has an approximately normal distribution with mean μx = 80 and standard deviation σx = 52. (b) Find the z value corresponding to x = 86.5. z = (c) Find P(x < 86.5). (Round your answer to four decimal places.) P(x < 86.5) = (d) Would it be unusual for a random sample of size 64 from the x distribution to have a sample mean less than 86.5? Explain. Yes, it would be unusual because less than 5% of all such samples have means less than 86.5. No, it would not be unusual because less than 5% of all such samples have means less than 86.5. No, it would not be unusual because more than 5% of all such samples have means less than 86.5. Yes, it would be unusual because more than 5% of all such samples have means less than 86.5.

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Transcript

00:01 There is given a normal distribution here and the mean and the standard deviation was given.

00:05 So the mean, which is denoted by new, which is 80 and the standard deviation, denoted by sigma, that was given as 52 here.

00:15 So i can define the random variable x, which is normally distributed, 80 and 52.

00:21 But there is a random sample size here.

00:24 So the n was given, which is 64.

00:27 So what we have to do, describe the x distribution of the mean and standard division of the distribution.

00:34 So the mean of the distribution here, which is for the sample, the distribution should be normal because the n is equal to 64, which is greater than 30.

00:45 We can say this is approximately normal.

00:50 And the sample mean, so the sample mean, which is equal to the population mean, which is 80, and the sample standard deviation.

01:00 So the sample standard deviation, which is equal to population, standard division divided by square root of the sample size, which is n here.

01:08 So let me just note the sample standard division as sigma x, which is 52 divided by square root of 64, which is equal to 52 over 8.

01:20 That would be, this is 52 divided by 8, which is equal to 6 .5.

01:25 So i can define the random variable x bar, which is normally distributed.

01:29 So the mean is 80 for the sample and the standard division is 6 .5 here.

01:35 And what about the z value here? so the z value for x is equal to and the corresponding x value.

01:46 So the x is equal to 86 .5 and the z score, which is equal to x minus mu over standard division.

01:53 So the z score for 86 .5 minus the mean is 80 and the standard division, which is 52.

02:00 That would be this is 6 .5 divided by 52 which is equal as 0 .125.

02:08 This is 0 .125 is the z score for 86 .5 and we have to get the probability of random variable x, which is less than.

02:17 This is 86 .5, which is equal to the z is less than 0 .125...

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