Deliverable 4 - Hypothesis Tests

Running Head: APPLICATION OF NORMAL DISTRIBUTION 1

APPLICATION OF NORMAL DISTRIBUTION 5

Application of Normal Distribution

Student’s Name

Institution Affiliation

1<p>

The total area under the normal curve with mean zero and standard deviation one is equal to one; Mean =0; Std. = The probability range is calculated as follows P (-1.53>X<1.98) where the Z-scores are found using the following formula; Z1=(x-mean)/std = (-1.53-0)/1 = -1.53 and Z2= (1.98-0)/1. Using left hand distribution table in the excel P (Z2<1.98) = 0.9761 and P (Z1<-1.53) =0.0630.

The second step is to find the difference between the probabilities scores where the smallest probability score is subtracted from the largest probability score derived from the z-scores in the standard normal distribution.

Specifically, 0.0630 is subtracted from the 0.9761 to arrive at a probability of the selected subject to 0.9131. Mathematical subtraction is carried out in this case because both z-scores are calculated from the left, and there is a region included when calculating the probability of each score which is not required and the only way to exclude it is through subtraction.

2. <b>

Step one is to state the mean and mean of the standard normal distribution, where mean=0 and std. =1

Step two is to convert the height interval to the standard normal distribution so that so that the z-scores can be calculated and be used to calculate the probability that will be necessary to calculate the percentage of the women that meet the requirement. The following relationship is used to do standardization; z=(x-µ)/std., (Matthew, 2013), where mean is 65 and STD is 3.5.

Specifically, probability P (64<X<77) will be standardized using above formula to appear as follows; for 64, Z= (64-65)/3.5=-2.86 and 77, Z= (77-6the 5)/3.5=3.428 and the probability range is as follows, P (-2.86<Z<3.43).

Using the excel spreadsheet, the probability of the z-scores are as follows, for P (Z=-2.86) =0.387548481 and the P (Z=3.43) is 0.999696617.

The last step is getting the difference between the two where (0.999696617-0.38754848) multiplied by 100 to give the percentage of the women that meet the height requirement as 61.2148135%.

3. <i>

Step one is writing them down the formula for standardization which is, Z=(x-u)/STD. In this particular question, the student was right in writing down the formula. However, the student interchanged the values in the numerator thus leading to a wrong answer. In this scenario, X=66, mean= 69.4 and std. =11.3, where if substituted into equation (66-69.4)/11.3 = -0.300884956 (Z-score).

4<b>.

The first step is to split -0.876 into -0.8 and -0.076 and round it off. Then in the z-table, look for 0.8 in the left column and the 0.08 which has been rounded off in the top row. Since the z-score is negative in this case, the area is arrived at by subtracting the z-value found from 0.500.

5<i>

The answer given by the student was wrong because plugged in 0.6573 into excel without considering the already the area is more than 50% and was from the right. In this scenario, the student first is required to subtract 0.6573 from one. Step two is taking the difference and plugging it into the inverse of normal distribution function to obtain the z-score for 0.6573 from right, NORM.INV (0.3427, 0, 1), which gives the z-score as -0.41.

6<p>

The probability that men will fit into the man-hole is the question to be answered in this question. The first step is to find the probability that the men’s shoulder is less than 22.5 inches, using the following formula, p(x<22.5). Step two would be standardizing the value of X to get to score, Z= (22.5-18.2)/2.09= 2.057416268. Standardization is crucial to give a value that a z-score that can be looked up for in the standard distribution table to obtain probability. The third step would look up for z-score in the table but in this case, have used excel spreadsheet to calculate the probability which is equal to 0.980176899. The last step is answering the question by getting the percentage of the men that will fit in the diameter which is got by (1- to 0.980176899) multiplied by 100 to give 1.9823101%

Reference

Matthew S.(2013). How to find a percentage from standard deviation and mean? Retrieved from, https://www.wyzant.com/resources/answers/8703/how_to_find_a_percentage_from_standard_deviation_and_mean