Statistics

What are the primary advantages of the Standard Normal Distribution that other normal distributions do not have?

The standard normal distribution normalizes the data by changing all the raw data points that tells us how many standard deviations the raw data points are away from the mean.

Find the standard z-score that corresponds to the following: Eighty percent of the distribution is below (to the left of) this value.

P (z<0.845) = .8. So the z-score that corresponds to 80% of the distribution below a random variable X is 0.845

Response

From my view, one of the greatest merit of standard normal distribution is the fact that tables are presented in any statistics textbook for the area under the curve for the normal curve and they are thus essential to answer questions about the original distribution. No naturally measured consists of this distribution. However, all other normal distributions are equivalent to this distribution when the unit of measurement is changed to measure standard deviations from the mean. Also, normal distribution can be converted in standard form in a manner in such that standardized tables can be used to tally the results.

The z-score will be obtained by:

Z score at 0.7995 = 0.84

Z score at 0.8023= 0.85

Z score at (0.84+0.85)/2 = 1.69

Z (0.8) = 1.69/2

Z(0.8) = 0.845

To find the z-score we find a near value to 0.8 which in this case is 0.7995 and 0.8023. To find the z-score of 0.8, we add the z-score of both probabilities and divide by 2. The z-score hence corresponds to 80% of the distribution below a random variable X is 0.845.

2. Discussion question with solutions

(1.) The standard normal probability density function is symmetric with respect to its mean of zero. From this it follows immediately that for any real number a,
P (z < -a) = P (z > a). (Note that the condition a > 0 is not necessary for this answer to be true, but clearly if the result is true for all real a, then it must be true for all positive real a.)

Response

When it comes to computing probabilities, probability density functions are indispensable. It is true to say that a>0 is not a necessary condition.

(2.) Since the area in the left tail of the standard normal distribution is less than 0.5, this problem can be expressed as (z) = 0.32. Once you recognize this, then finding z is done by either plugging a number into a calculator or looking up the corresponding number in a table of standard normal values.

Therefore: z = -0.4677

Response

The probability of scores from the statistical tables begin from 0.5 and since 0.32 is below 0.5 it implies that the score falls to the left area of the standard distribution. To compute the z-score of the figure, we will subtract it from 1 to obtain 0.68 (1-0.32). Locating the z-score from the statistical tables, we obtain the z-score to be (0.46+0.47)/2 which is equal to 0.93/2 = 0.465. The negative sign has to be put to show direction and the fact that its area falls to the left of the standard normal distribution.