A fruit packing company produced peaches last summer whose...

A fruit-packing company produced peaches last summer whose weights were normally distributed with mean 14 ounces and standard deviation 0.4 ounce. Among a sample of 1000 of those peaches, about how many could be expected to have weights between 14.6 and 15 ounces? The number of peaches expected to have weights of at least between 14.6 and 15 ounces? _____ (ROUND to the NEAREST WHOLE NUMBER)

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00:01 A fruit packing company produces these peaches, and they are normally distributed.

00:07 So let's draw their distribution.

00:12 The mean, mu, is 14.

00:15 Standard deviation, sigma, is 0 .4.

00:18 We want to know that if we were to take 1 ,000 of these peaches, how many would we expect to have weights between 14 .6 and 15 ounces? so 14 .6 is above the mean.

00:31 So is 15.

00:32 What percentage of peaches would fall within this little area here? well, let's find out.

00:43 First of all, we can't use raw data with a normal distribution that we have to convert it into z scores.

00:48 And z equals x minus mu over sigma.

00:52 So the z score for 15 and the z score for 14 .6.

01:00 Okay, so 15 minus 14 divided by 0 .4 is 2 .5.

01:08 1 .6 minus 14 divided by 0 .4 is 1 .5.

01:15 So what we're saying is 14 .6 is 1 .5 standard deviations above the mean.

01:20 15 is 2 .5 standard deviations above the mean.

01:24 And we can take these z scores and we can find probabilities associated with them.

01:29 You fill out this little area here in red.

01:33 If you go to a z score table and you put in the value 2 .5, you will get the distanced mean, which is red plus green.

01:41 If you put in 1 .5, again, you will get the distance to mean, which will just be this red area here...

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