GATE BT 2016 • Q32 NAT • 1 Mark

Section 1: Engineering Mathematics

Probability and Statistics

Subtopic: Discrete & Continuous Probability Distributions

Question Statement:

A variety of genetic elements are used in the transgenic plant research. Match the genetic elements (Column-I) with their corresponding source (Column-II). Column-I Column-II P. Ubiquitin1 promoter 1. Agrobacterium tumefaciens Q. Nos transcriptional terminator 2. Streptomyces hygroscopicus R. bar selection marker gene 3. Escherichia coli S. gus reporter gene 4. Zea mays (A) P-2, Q-1, R-3, S-4 (B) P-2, Q-3, R-4, S-1 (C) P-3, Q-4, R-1, S-2 (D) P-4, Q-1, R-2, S-3

Official Answer Key:

24.5 - 28.5

Step-by-Step Solution & Explanation

### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2016 Q.32)** **1. Governing Concept & Definition**: - **Topic**: Statistics & Sample Standard Deviation ($s$) - For a sample of $n$ observations $\{x_1, x_2, \dots, x_n\}$, the sample mean $\bar{x}$ and sample standard deviation $s$ are: $\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i, \quad s = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n - 1}}$ **2. Detailed Step-by-Step Calculation**: - **Given Runs**: $55, 75, 67, 88, 15$ ($n = 5$) - **Mean ($\bar{x}$)**: $\bar{x} = \frac{55 + 75 + 67 + 88 + 15}{5} = \frac{300}{5} = 60$ - **Deviations and Squares**: - $(55 - 60)^2 = (-5)^2 = 25$ - $(75 - 60)^2 = (15)^2 = 225$ - $(67 - 60)^2 = (7)^2 = 49$ - $(88 - 60)^2 = (28)^2 = 784$ - $(15 - 60)^2 = (-45)^2 = 2025$ - **Sum of Squares**: $\sum (x_i - \bar{x})^2 = 25 + 225 + 49 + 784 + 2025 = 3108$ - **Standard Deviation**: - Sample standard deviation ($n - 1 = 4$): $s = \sqrt{\frac{3108}{4}} = \sqrt{777} \approx 27.87$ - Population standard deviation ($n = 5$): $\sigma = \sqrt{\frac{3108}{5}} = \sqrt{621.6} \approx 24.93$ **3. Final Answer**: - **Calculated Value**: **27.87** (or **24.93**) - **Official GATE Accepted Range**: **24.5 to 28.5**

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