A survey found that women's heights are normally distributed with mean 63.5 in. and standard deviation 2.3 in. The survey also found that men's heights are normally distributed with a mean 68.7 in. and a standard deviation 2.8.
(a) Most of the live characters at an amusement park have height requirements with a minimum of 4ft 9in and a maximum of 6ft 9in.
Find the percentage of women meeting the height mean 68.7 in and standard deviation 2.8.
http://davidmlane.com/hyperstat/z_table.html
To find the percentage of women meeting the height requirement of mean 68.7 in and standard deviation 2.8, we need to calculate the area under the normal distribution curve.
First, let's standardize the height requirement using the formula:
Z = (X - μ) / σ
Where:
Z = the standard score (also known as Z-score)
X = the height requirement (68.7 in)
μ = the mean height of women (63.5 in)
σ = the standard deviation of women's heights (2.3 in)
Plugging in the values:
Z = (68.7 - 63.5) / 2.3
Calculating Z:
Z = 2.260869565217391
Now, we need to find the corresponding area under the standard normal distribution curve to determine the percentage.
You can use a standard normal distribution table or a calculator to find the area. Let's use a calculator to find the percentage.
Using a calculator's normal distribution function, we can find the percentage of women meeting the height requirement.
Assuming a normal distribution, we can use the cumulative distribution function (CDF) to find this percentage. We want to find the area to the right of the Z-score since we are interested in the percentage of women meeting a height requirement greater than 68.7 inches.
Using a calculator:
1. Go to the normal distribution function on your calculator.
2. Enter the Z-score of 2.260869565217391.
3. Set the lower bound to 2.260869565217391 and the upper bound to positive infinity.
4. Calculate the CDF, which represents the area to the right of the Z-score.
Once you obtain the result, you will have the percentage of women meeting the height requirement of 68.7 inches and a standard deviation of 2.8 inches.
To find the percentage of women meeting the height mean of 68.7 inches and standard deviation of 2.8 inches, we need to calculate the z-score and use a z-table.
The z-score formula is:
z = (x - μ) / σ
Where:
x = value we want to find the percentage for (68.7 inches)
μ = mean (63.5 inches)
σ = standard deviation (2.3 inches)
Substituting the values:
z = (68.7 - 63.5) / 2.3
z ≈ 2.2609
Next, we need to find the percentage using the z-table. The z-table provides the probabilities associated with specific z-scores. We need to find the probability corresponding to z = 2.2609.
Looking up the z-score in the z-table, we find that the probability associated with a z-score of 2.2609 is approximately 0.9877.
To convert this into a percentage, we multiply the probability by 100:
0.9877 * 100 ≈ 98.77%
Therefore, approximately 98.77% of women meet the height requirement with a mean of 68.7 inches and standard deviation 2.8 inches.