Find the probability that a part selected at random would have a length It was found that the mean length of `100` parts produced by a lathe was `20.05\ "mm"` with a standard deviation of `0.02\ "mm"`. We find the area on the left side from `z = -1.06` to `z = 0` (which is the same as the area from `z = 0` to `z = 1.06`), then add the area between `z = 0` to `z = 4.00` (on the right side): (d) This is the same as asking "What is the area between `z=-1.06` and `z=4.00` under the standard normal curve?" (c) This is the same as asking "What is the area between `z=1.06` and `z=4.00` under the standard normal curve?" This time, we need to take the area of the whole left side (`0.5`) and subtract the area from `z = 0` to `z = 2.15` (which is actually on the right side, but the z-table is assuming it is the right hand side.) (b)This is the same as asking "What is the area to the left of `-2.15` under the standard normal curve?" We need to take the whole of the right hand side (area `0.5`) and subtract the area from `z = 0` to `z = 1.06`, which we get from the z-table. (a)This is the same as asking "What is the area to the right of `1.06` under the standard normal curve?" Example 3įind the area under the standard normal curve for the following, using the z-table. We can also use Scientific Notebook, as we shall see. From this table the area under the standard normal curve between any two ordinates can be found by using the symmetry of the curve about z = 0. The areas under the curve bounded by the ordinates z = 0 and any positive value of z are found in the z -Table. This means that `68.27%` of the scores lie within `1` standard deviation of the mean. In the above graph, we have indicated the areas between the regions as follows: Standard Normal Curve showing percentages μ = 0, σ = 1. This area is graphed as follows:ġ 2 3 0 -1 -2 -3 Z f(Z) 68.27% 95.45% 99.73% Open image in a new page One-half standard deviation = `σ/2 = 1/6`, and P( x 1 < X < x 2) = P( z 1 < Z < z 2) Example 2Ĭonsidering our example above where `μ = 2`, `σ = 1/3`, then Hence, we have the following equivalent probabilities: The area under the Z curve between Z = z 1 and Z = z 2. The area under the X curve between X = x 1 and X = x 2 Have corresponding Z values between z 1 and z 2, it means: Since all the values of X falling between x 1 and x 2 If we have mean μ and standard deviation σ, then Formula for the Standardized Normal Distribution Standardizing the distribution like this makes it much easier to calculate probabilities. The new distribution of the normal random variable Z with mean `0` and variance `1` (or standard deviation `1`) is called a standard normal distribution. The two graphs have different μ and σ, but have the same area. Standard Normal Curve μ = 0, σ = 1, with previous normal curve If we have the standardized situation of μ = 0 and σ = 1, then we have:ġ 2 3 -1 -2 -3 0.5 1 Z Open image in a new page It makes life a lot easier for us if we standardize our normal curve, with a mean of zero and a standard deviation of 1 unit. The probability of a continuous normal variable X found in a particular interval is the area under the curve bounded by `x = a` and `x = b` and is given byĪnd the area depends upon the values of μ and σ. Area Under the Normal Curve using Integration In a normal distribution, only 2 parameters are needed, namely μ and σ 2. It is completely determined by its mean and standard deviation σ (or variance σ 2) The total area under the curve is equal to 1 The mean is at the middle and divides the area into halves The normal curve is symmetrical about the mean μ Continues below ⇩ Properties of a Normal Distribution
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