Leibniz's calculator
You will need to use tables of z-score or a z-score calculator. You cannot derive the value analytically.The required z-score is 0.524401
You either look it up in a table of z scores or you can use a calculator such as the TI8 and use normalcdf.
The average z score chart lists z scores with three significant figures. For example, you can find the z score -1.81 on the chart, but not -1.812 or -1.818. In the case that you wish to look up a z score with more than three significant figures, round it to three significant figures and then use the chart. OR You can also use a calculator if you wish to get more accurate results. The link for calculator is mentioned below.
Oh, what a happy little question! To put arcsec in a calculator, you simply press the "2nd" or "Shift" key on your calculator, then find the "sec" button. This will allow you to calculate the arcsec of an angle and create beautiful mathematical landscapes. Just remember, there are no mistakes, only happy little calculations!
7x + 30xz I use the symbolab calculator to solve math equations I can't figure out :)
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.
In BASIC, this could be as simple as: 10 Input X 20 Input Y 30 Z=X+Y 40 Print X;" + ";Y;" = ";Z 50 END In JAVA... /** * A simple calculator that adds, subtracts, multiplies, and divides. * Written in Java by: AustinDoggie */ public class Calculator() { public Calculator() { // don't really have to do anything here } public int add(int x, int y) { int z = x + y; // add two numbers return z; // and return it } public int subtract(int x, int y) { int z = x - y; // subtract two number return z; // and return it } public int multiply(int x, int y) { int z = x*y; // multiply two number return z; // and return it } public int divide(int x, int y) { int z = x/y; // divide two numbers return z; // and return the result } } Hope that helps.
To find the value of ( z ) such that the area between (-z) and (z) under the standard normal distribution curve is 0.777, we need to determine the corresponding cumulative probability. The area between (-z) and (z) represents the central portion of the distribution, so we can express this as: ( P(-z < X < z) = 0.777 ). This means that ( P(X < z) - P(X < -z) = 0.777 ). Since the distribution is symmetric, ( P(X < -z) = 1 - P(X < z) ), leading to ( 2P(X < z) - 1 = 0.777 ) or ( P(X < z) = 0.8885 ). Using a standard normal distribution table or calculator, we find that ( z \approx 1.175 ).
P(0 < z < 2.53) refers to the probability that a standard normal random variable (z) falls between 0 and 2.53. To find this probability, you would typically look up the z-scores in a standard normal distribution table or use a calculator. The cumulative probability for z = 2.53 is approximately 0.994, and for z = 0, it is 0.5. Therefore, P(0 < z < 2.53) is approximately 0.994 - 0.5 = 0.494.
To find the z-score where 8% of the distribution's area lies between -z and z, we first recognize that this means 4% (or 0.04) lies in each tail of the normal distribution. Therefore, we need to find the z-score that corresponds to the cumulative area of 0.04 in the left tail. Using standard normal distribution tables or a calculator, we find that the z-score for 0.04 is approximately -1.75. Thus, the positive z-score is approximately 1.75, meaning z ≈ 1.75.
To find the area under the normal distribution curve between z = 1.52 and z = 2.43, you can look up the z-scores in a standard normal distribution table or use a calculator. The area to the left of z = 1.52 is approximately 0.9357, and the area to the left of z = 2.43 is approximately 0.9925. Subtracting these values gives an area of approximately 0.9925 - 0.9357 = 0.0568. Thus, the area between z = 1.52 and z = 2.43 is about 0.0568.
To find the proportion of a normal distribution corresponding to z-scores greater than +1.04, you can use the standard normal distribution table or a calculator. The area to the left of z = 1.04 is approximately 0.8508. Therefore, the proportion of the distribution that corresponds to z-scores greater than +1.04 is 1 - 0.8508, which is approximately 0.1492, or 14.92%.