Suppose that the mean value of interpupillary distance (the...

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. (a) If the distribution of interpupillary distance is normal and a random sample of n = 25 adult males is to be selected, what is the probability that the sample mean distance x for these 25 will be between 63 and 67 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.) P = 0.8186 What is the probability that the sample mean distance x for these 25 will be at least 68 mm? P = 0.0228 (b) Suppose that a sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 63 and 67 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.) P = 0.9772

Transcript

00:01 The mean and the standard deviation were given in the question here.

00:05 So the mean value denoted by this is mu, and that was given as 65 millimeter.

00:10 And the standard division was given also.

00:13 Standard division denoted by sigma, and that was given as 5mm here.

00:17 Let me just take a look at the part a.

00:20 So in this case, the distribution of this, which is normal, and the random sample of n is equal to 25.

00:26 So the sample size was also given here.

00:28 So if the distribution is normal, so we have to find the sample mean.

00:34 So the sample mean is equal to the population mean, which is 65.

00:38 And the sample standard division, so the sample standard division for this one, i'm going to just apply the central dimith theorem, which says the population standard division divided by square of the sample size.

00:49 Let me denote this is x bar, which is equal to this is 5, and divided by square root of 25, which is equal to 5 over 5, which is 1.

00:59 Can define random variable x bar for the sample, which is normally distributed.

01:03 So the mean is 65 and the standard division, which is one here.

01:07 So what we need to get? so we need to get the probability of this random variable x bar that is between 63 and 67.

01:16 So in order to get this one, i'm going to use the graphic display calculator application, the normal cdf, so the lower boundary is 63, upper boundary is 67, and the mean is 65 and the standard division which is one here just press second variance the second option here the lower boundary is 63 and the upper boundary which is 67 so the mean is 65 and the standard division one so the probability would be which is um zero point this is 95 and 45 with four decimal places and what about for the second part of the choice a which is again for this sample we need to get the probability of the random variable x bar, which is at least means this is greater than or equal to 68.

02:00 So again, i'm going to use the normal cdf.

02:04 Lower boundary is 68.

02:05 There is no upper boundary.

02:06 I'm going to put a very big number.

02:07 So the mean is 65 and the standard division 1...

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