The histogram shows the distribution of the diameters of 100...

The histogram shows the distribution of the diameters of 100 rainbow trout eggs. The distribution is approximately normal with a mean of about 4.31 mm and standard deviation of about 0.25 mm. Frequency Rainbow Trout 20 Mean \approx 4.31 mm 16 SD \approx 0.25 mm 12 8 4 0 3.7 3.9 4.1 4.3 4.5 4.7 4.9 Egg diameter (mm) a) Approximately what proportion of the diameters lie within 2 standard deviations of the mean? 4.81 X b) Approximately what proportion of the diameters lie within 3 standard deviations of the mean? 3.56 X

Transcript

00:01 All right, so we have a large tank of fish from the hatchery is being delivered to a lake.

00:05 The hatchery claims the mean length of the fish is 15 inches, and the standard deviation is 3 inches here.

00:16 And then we take a sample of 53 fish, and the samples from the tank, and we want x to be the mean sample length of these fish.

00:28 What's the probability of the x is within half an inch of the claimed to popperature? population mean.

00:33 So the way i tackled this is i looked at the margin of error of our confidence interval.

00:39 So normally we'd have something like this x bar plus minus z alpha over two multiplied by the standard error, which is the standard deviation divided by the square of the sample size.

00:53 And this term here is called a margin of error.

00:56 And what we want is the margin of error.

01:02 We want this term to be less than half an inch.

01:08 So i'll do the calculations over here.

01:10 So 0 .5.

01:12 We want this whole thing, z.

01:14 I'm just going to write z for now, sigma over root n.

01:18 We want this whole thing to be less than half an inch.

01:21 And we have sigma and we have square of n.

01:22 So we just find z.

01:23 And then we're going to find z because if we find z, then we can find the probability.

01:28 So it's going to do that.

01:30 So essentially we're going to do is take 0 .5 divided by this term.

01:35 So 0 .5 .5.

01:36 So 0 .5 .5.

01:36 5 divide by sigma over root n.

01:42 Sigma is 3.

01:44 N is 53.

01:48 And this can be less than z.

01:49 And that gives us a z score of this, 1 .121.

01:55 So that's going to give us, if here's the z score of zero, we'll say this is 1 .21.

02:06 What we can do is to do a little lookup to look up to look up the z score in a, textbook z table or i use my spreadsheet and i found the air to left of it to be 0 .8875.

02:18 But that's not what we want because what that means is that's going to give us this.

02:27 Oh, and by the way, this is normal because our sample size is greater than 30.

02:31 I should say that.

02:32 That's why we have the normal distribution here.

02:34 All right, back to the graph here.

02:36 So it's 0 .887.

02:37 That's all this stuff.

02:38 But remember, we don't want that.

02:39 We want to be contained within a value.

02:42 So we want to be within the negative 1 .21.

02:48 So we want to be within this chunk here.

02:51 So we want all this stuff...

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