A z table is used to calculate the probability of choosing something that is normally distributed. In order to use it, first a z score is needed. A z score is the number of standard distributions a value is away from the mean of the data. In order to find the z score, take the value of the datum, subtract the mean, then divide by the standard deviation. The result is a z score. Look up the z score on the table to find the probability of getting anything equal to or lesser than the value you chose.
See the attached link. The area that is read from the table is 1 - .9066 or .0934. Go to the body of the table for the value .0934 and the answer is -1.32. Therefore Z = -1.32.
Since the normal distribution is symmetric, the area between -z and 0 must be the same as the area between 0 and z. Using this fact, you can simplify this problem to finding a z such that the area between 0 and z is .754/2=.377. If you look this value up in a z-table or use the invNorm on a calculator, you will find that the required value of z will be 1.16. Therefore, the area between -1.16 and 1.16 must be approximately .754.
1.555 With 88% confidence, there is 6% (0.06) in either tail of the standard Normal distribution. Table C will not help here. Using Table A the correct z* is about halfway between 1.55 and 1.56. According to technology, z*=1.555
You can use a table or a graph to organize you findings.
From the table in the related link, the value at z equal one is 0.3413. The area then to the right of z equal one is 0.5 - 0.3413, or 0.1587.
You either look it up in a table of z scores or you can use a calculator such as the TI8 and use normalcdf.
If the sample size is less then 30 use the T table, if greater then 30 use the Z table.
Not all z-score tables are the same. You must know how to use the specific table that you have.
For statistical tests based on (Student's) t-distribution you use the t-table. This is appropriate for small sample sizes - up to around 30. For larger samples (or degrees of freedom), the t-distribution becomes very close to the Standard Normal distribution so you use the z-tables.
Not a singular "Z", but there are two: Zn- Zinc Zr- Zirconium.
YES
When we use a z-test, we know the population mean and standard deviation. When we use a t-test, we do not know the population standard deviation and thus must estimate this using the sample data that we have collected. If you look at your z-table and t-table, tcrit for df(infinity) = zcrit because at df(infinity) we would have an entire population and no longer need an estimate.
See the attached link. The area that is read from the table is 1 - .9066 or .0934. Go to the body of the table for the value .0934 and the answer is -1.32. Therefore Z = -1.32.
It is not easy to illustrate this in a table since it would need to be a 3-d table. The browser which we are required to use is bad enough for ordinary tables. So here goes:X = True, Y = True, Z = True then (X or Y or Z) = TrueX = True, Y = True, Z = False then (X or Y or Z) = TrueX = True, Y = False, Z = True then (X or Y or Z) = TrueX = True, Y = False, Z = False then (X or Y or Z) = TrueX = True, Y = True, Z = True then (X or Y or Z) = TrueX = True, Y = True, Z = False then (X or Y or Z) = TrueX = True, Y = False, Z = True then (X or Y or Z) = TrueX = True, Y = False, Z = False then (X or Y or Z) = FalseThe simple way to remember is that the OR gate gives False only when each input is False.
Mendelevium
If the value given in the table for Z = z is k: that is, pr(Z > z) is 1 - k, then the two-tailed probability of observing a value which is at least as extreme, ie Pr(|Z| > z) is 0.5*(1-k).
P(0 < Z < z) = 0.3770 is the same as looking at P(Z < z) = 0.8770 because the other half of the curve (anything less than 0) has probability of 0.5. Now this is a problem of just looking it up from the table. The table gives a value z = 1.16 for the probability of 0.8770. So P(0 < Z < 1.16) = 0.3770.