Central Tendency & DispersionMTP March 21Question 3170 of 473
All Questions

What is the standard deviation of number recoveries among 48\displaystyle 48 patients when the probability of recovering is 0.75\displaystyle 0.75?

Options

A36\displaystyle 36
B81\displaystyle 81
C9\displaystyle 9
D3\displaystyle 3
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option d — 3\displaystyle 3

All Options:

  • A36\displaystyle 36
  • B81\displaystyle 81
  • C9\displaystyle 9
  • D3\displaystyle 3

Detailed Solution & Explanation

**Given:** n=48\displaystyle n = 48 patients, p=0.75\displaystyle p = 0.75 (probability of recovery), q=1−p=0.25\displaystyle q = 1 - p = 0.25 This is a Binomial distribution problem. **For Binomial Distribution:** Mean=np=48×0.75=36\text{Mean} = np = 48 \times 0.75 = 36 Variance=npq=48×0.75×0.25=9\text{Variance} = npq = 48 \times 0.75 \times 0.25 = 9 Standard Deviation=npq=9=3\text{Standard Deviation} = \sqrt{npq} = \sqrt{9} = 3 The SD = **3**, which is Option D. Option A (36) is the **Mean**, not the SD. The given `correct_option` 'a' (36) is the mean, not SD. The correct SD = 3. Hence, **Option D** is the correct answer.

More Questions from Central Tendency & Dispersion

If the range of 28,22,40,20,15,50\displaystyle 28, 22, 40, 20, 15, 50 is

If the S.D. of the 1st\displaystyle 1^{st} n\displaystyle n natural numbers is 30\displaystyle \sqrt{30} then the value of n\displaystyle n is

If the variance of 5,7,9\displaystyle 5, 7, 9 and 11\displaystyle 11 is 4\displaystyle 4, then the coefficient of variation is:

A helicopter flies from A to B at the rate of 500 km/hr. and comes back at the rate 700 km/hr. The average speed of the helicopter is

If Arithmetic Mean (A.M.) and Geometric Mean (G.M.) of two numbers are 6.50 and 6 respectively, then the two numbers are

Which of the following is not a method of dispersion?

Ready to Master Central Tendency & Dispersion?

Practice all 473 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free