Suppose the sediment density gcm of a randomly selected...

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.63 and standard deviation 0.76. (a) If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between 2.63 and 3.00? (Round your answers to four decimal places.) at most 3.00 between 2.63 and 3.00 (b) How large a sample size would be required to ensure that the first probability in part (a) is at least 0.99? (Round your answer up to the nearest whole number.) specimens

Transcript

00:01 There is a normal distribution here and the given mean value, which is denoted by mu, that is 2 .63, and the standard division, which is given as 0 .76.

00:12 Let's say x be the random variable, but we don't need the x here because if a random sample of 25 specimens.

00:20 So n is given as 25 for the first part.

00:24 So what we have to do, we have to find the standard division for the sample.

00:28 It has the formula which is standard deviation of the sample i mean the population and divided by the square two of the sample size so from this information the sample standard division which is 0 .76 divided by radical 25 0 .76 divided by 5 which is 0 .76 divided by 5 which gives us 0 .152 let's say x bar the random variable for this normal distribution the mean value.

00:59 Is the same with the population 2 .63 and the standard division 0 .152.

01:05 So the first one, the question is the probability that the sample average sediment density is at most 3.

01:13 So what that means the probability we need to find x is x bar is less than or equal to 3.

01:22 So what we have to do, let me just graph this situation.

01:26 This is the normal graph.

01:28 So the mean value is 2 .63.

01:31 And the 3 is here, so we need to get the area of this region.

01:35 In order to get this one, we can use the normal cdf function.

01:40 So the lower boundary that goes to negative infinity, instead of negative infinity, we are going to put a very small number, which is negative 1 -a -99, and comma, the upper boundary here is 3, and the mean value is 2 .63 in the standard deviation, which is 0 .152.

01:59 Let's get the answer.

02:00 Second distribution normal cdf lower boundary negative 1 this is second e99 comma and upper boundary is 3 then 2 .63 comma 0 .152 we got the probability 0 .99 25 so this is the answer for the first part and what about the probability between 2 .63 and 3 so we have to get the argue of this region here so let me just use another color.

02:34 So that we want the x bar is between 3 and 2 .63.

02:38 Again, we can use the normal cdf function.

02:42 And the lower boundary 2 .63, upper boundary 3...

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