Theoretical DistributionUsing the theoretical distribution, complete the following statements. Use a normalapproximation based on the sample mean and standard deviation.IQR = _______The 15 percentile is _______.The 85 percentile is _______.Median is _______. Attached I have the data I am currently working with and the assignment. I need help with the Theoretical Distribution section above. ### Calculate the Following3. Calculate the following. - a. \( \bar{x} = \_\_\_\_\_ \) - b. \( s = \_\_\_\_\_ \)4. Draw a smooth curve through the top of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it have a v-shape, does it have a hump in the middle or at either end, and so on?)### Analyze the DistributionUsing your sample mean, sample standard deviation, and histogram, what was the approximate theoretical distribution of the data you collected?- \( X \sim \_\_\_ \left( \_\_\_\_\_ , \_\_\_\_\_\_ \right) \)- How does the histogram help you arrive at the approximate distribution?### Describe the DataUsing the data you collected complete the following statements. (Hint: order the data)#### Remember\( \text{IQR} = Q_3 - Q_1 \)- **IQR = \_\_\_\_\_**- The **15th percentile** is \_\_\_\_\_.- The **85th percentile** is \_\_\_\_\_.- **Median** is \_\_\_\_\_.- What is the theoretical probability that a randomly chosen pinky length is more than 6.5 cm?- Explain the meaning of the **85th percentile** of this data.### Theoretical DistributionUsing the theoretical distribution, complete the following statements. Use a normal approximation based on the sample mean and standard deviation.- **IQR = \_\_\_\_\_**- The **15th percentile** is \_\_\_\_\_.- The **85th percentile** is \_\_\_\_\_.- **Median** is \_\_\_\_\_. ### Data on Pinky Length#### Table of Pinky Length Measurements- Measurements: 7, 7, 7, 7, 6.5, 6.5, 6.5, 6.5, 6, 6, 7, 7, 5, 6.5, 7, 6.5, 6, 6.5, 6.5, 6, 5.5, 6.5, 6, 6.5, 6.5, 6, 6, 5.5, 6, 6.5#### Statistical Summary- **Average (Mean):** 6.433333- **Standard Deviation (S):** 0.486602- **1st Quartile (Q1):** 6- **2nd Quartile (Q2/Median):** 6.5- **Interquartile Range (IQR):** 0.875- **15th Percentile:** 6- **85th Percentile:** 7- **Normal Distribution Approximation:** - \( X \sim \mathcal{N}(6.43, 0.487) \) #### Probability Calculation- **Probability of Length Greater than 6.5 cm:** - \( P(X > 6.5) = 0.266667 \)#### Histogram DescriptionThe histogram visualizes the frequency distribution of pinky lengths:- **Intervals:** 1. [5, 5.55] 2. (5.55, 6.1] 3. (6.1, 6.65] 4. (6.65, 7.2] - **Frequency:** - [5, 5.55]: 2 - (5.55, 6.1]: 6 - (6.1, 6.65]: 13 (most frequent) - (6.65, 7.2]: 9This data and its analysis allow us to understand the distribution of pinky lengths in the sample, highlighting the central tendency, variability, and the probability of certain lengths.

Given that: Pinky Length: 7, 7, 7, 6.5, 6.5, 6.5, 6.5, 6, 7, 7, 7, 5, 6, 7, 6.5, 6.5, 6, 6.5, 6, 6.5, 6, 5.5, 6.5, 6, 6, 6.5, 6.5, 6.5, 7, 6.5Arrange the numbers in ascending order: 5, 5.5, 6, 6, 6, 6, 6, 6, 6, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 7, 7, 7, 7, 7, 7, 7, 7 Find the median: N=30 is even, so median is the average of the two middle numbers at position 15 and 16. Median=n2thterm+n2+1thterm2=302thterm+302+1thterm2=15thterm+16thterm2=6.5+6.52=6.5 Thus, the required median is 6.5. Find Q1 by finding the median for lower half of data(left of the median): 5, 5.5, 6, 6, 6, 6, 6, 6, 6, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5 N=15 is odd, so median is the middle number at position 8. Q1=n+12thterm=15+12thterm=8th term=6 Thus, the value of Q1 is 6.Find Q3 by finding the median for upper half of data(right of the median): 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 6.5, 7, 7, 7, 7, 7, 7, 7, 7 N=15 is odd, so median is the middle number at position 8. Q3=n+12thterm=15+12thterm=8th term=7 Compute the inter-quartile range or IQR by using the formula: IQR=Q3-Q1=7-6=1 Thus, the required IQR is 1. Compute the 15th percentile by using the formula: Here, n=30. P15=15n+1100th value of the observation=1530+1100th value of the observation=(4.65)th value of the observation =4th observation +0.65[5th-4th] =6+0.65[6-6] =6+0.65(0) =6+0 =6 Thus, the required 15th percentile is 6. Compute the 85th percentile by using the formula: Here, n=30. P85=85n+1100th value of the observation=8530+1100th value of the observation=(26.35)th value of the observation =26th observation +0.35[27th-26th] =7+0.357-7=7+0.350=7 Thus, the required 15th percentile is 7.