Question

10) Aditi has a rare disease due to which she has to take injection every third day. The cost of each injection is *20. Let
X represent her weekly bill. Find the standard deviation of her weekly bill (in rupees). (Correct upto 2 decimal points)​

Answer 1

Correct answer :Rs. 4.66

Steps : I am lazy

Answer 2

The standard deviation of her weekly bill = 14.14 Rs.

Step-by-step explanation:

Given:

Aditi has a rare disease for which she has to take an injection of Rs. 20 every third day

To find out:

The standard deviation of her weekly bill

Solution:

If Aditi takes the medicine at the first day of the week then she will taken the second injection on the 4th day and the third injection on the 7th day

Thus, the weekly bill = 20 × 3 = 60 Rs.

If she takes the injection on the second day of the week then she will take the second injection on 5th day

similarly if she takes the injection on the 3rd day of the week then she will take the second injection on 6th day and if she takes the injection on the 4th day of the week then she will take the second injection on 7th day

In these three cases her weekly bill = 20 × 2 = 40 Rs.

Mean of her weekly expanses

$\bar X=\frac{60+40}{2}$

$\implies \bar X=50$ Rs.

$N=2$

Therefore, the standard deviation

$\sigma=\sqrt{\frac{\Sigma (X-\bar X)^2}{N-1}}$

$\implies \sigma=\sqrt{\frac{(50-40)^2+(50-60)^2}{2-1}}$

$\implies\sigma=\sqrt{\frac{200}{2}}$

$\implies \sigma=10\sqrt{2}$

$\implies\sigma=10\times 1.414$

$\implies \sigma=14.14$

Hope this answer is helpful.

Know More:

Q: A sample of 900 members has a mean 3.4 cm and standard deviation 2.61 cm. Test whether the sample is from a large population of mean 3.25 cm and standard deviation 2.61 cm. If the  population is normal and its mean is unknown, find the 95% confidence interval for population  mean.

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