A bottling machine can be regulated so that it discharges an...

A bottling machine can be regulated so that it discharges an average of 𝜇 ounces per bottle. It has been observed that the amount of fill dispensed by the machine is normally distributed with 𝜎 = 1.0 ounce. If a sample of n = 9 filled bottles is randomly selected from the output of the machine on a given day (all bottled with the same machine setting), and the ounces of fill are measured for each, then the probability that the sample mean will be within 0.34 ounce of the true mean is 0.6923. Suppose that Y is to be computed using a sample of size n. (a) If n = 16, what is P(|Y − 𝜇| ≤ 0.34)? (b) Find P(|Y − 𝜇| ≤ 0.34) when Y is to be computed using samples of sizes: n = 25 P(|Y − 𝜇| ≤ 0.34) = n = 36 P(|Y − 𝜇| ≤ 0.34) = n = 49 P(|Y − 𝜇| ≤ 0.34) = n = 64 P(|Y − 𝜇| ≤ 0.34) =

Transcript

00:01 Hello everyone.

00:02 So in this question the value of sigma is given as 1 .0.

00:08 Now for the first part a given that n is equal to 16 we have to find the probability of modulus of y minus mu less than equals to 0 .34 so this is same as probability of y minus mu line between minus 0 .34 to 0 .34.

00:53 So this is same as.

00:55 Now to convert any random variable say y, which is following a normal distribution with mean mu and variant sigma square into standard normal variant, this standard normal variant z is equals to y minus mu divided by sigma whereas y bar follows normal distribution with mean mu and variance, sigma squared divided by n, and in this case, standard normal variant becomes z equals to y minus mu divided by sigma divided by square root n.

01:34 Right? so in this case, when we are converting the center term into standard normal variant, we already have numerator term.

01:44 So we'll simply divide it by sigma divided by square root n, that is probability of minus 0 .34 divided by sigma, that is 1, divided by square root.

01:55 The value of n is 16 less than equals to z less than equals to 0 .34 divided by 1 divided by square root 16.

02:08 So on solving this we get probability that z lies between minus 1 .36 to 1 .36.

02:21 Or this is same as 0 .8262.

02:30 So again coming to part b, we have to compute the probability given n equals to 25.

02:46 So we have to compute the probability mod y minus mu is less than equals to 0 .34.

02:57 So we will be proceeding in the same manner as above.

03:01 So this becomes probability minus 0 .34 less than equals to y minus mu, less than equals to 0 .34.

03:10 Again converting into standard normal variant, we have probability that z lies between minus 2 .04 and plus 2 .04.

03:26 How we have got this minus 0 .34, dividing by sigma, that is 1, divided by square root 25, which is equal to minus 2 .04.

03:40 So this is same as 2 times probability of z less than equals to 2 .04 minus 1, which is equals to 0 .9586.

04:00 Similarly, we have to calculate other probabilities, that is, at probability at n equals to 36.

04:11 So again, proceeding in the same manner, probability that modulus of y minus mu is less than equals to 0 .34...

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