Question

The weekly wages of 1000 workmen are normally distributed with a mean of 2000 and a standard deviation of 150. Estimate the number of workers whose wages lie between 1750 and 2250.

Answer 1

\(\mu\) = mean = 2000

\(\sigma\) = Standard deviation = 150

\(X \sim N(\mu, \sigma^2)\)

\(\frac{X - \mu}{\sigma} \sim N(0, 1)\)

⇒ \(\frac{X - 2000}{150} \sim N(0, 1)\)

\(\because 1750 \le X \le 2250\)

⇒ \(\frac {1750 - 2000}{150} \le \frac{X - 2000}{150} \le \frac{2210-2000}{150}\)

⇒ \(\frac{-250}{150} \le \frac{X - 2000}{150} \le \frac{250}{150}\)

⇒ \(\frac{-5}3 \le \frac {X - 2000}{150} \le \frac53\)

\(\because \frac{X - 2000}{150} \sim N(0, 1)\)

\(\therefore P\left(\frac{-5}3 \le \frac{X - 2000}{150} \le \frac 53\right) = P(Y \le \frac53) - P(Y \le \frac{-5}3 )\)

\(= 0.9516 - 0.0485\)

\(= 0.9031\)

The number of workers whose wages lie between 1750 & 2250 = 0.9031 x total no. of workers 

\(= 0.9031 \times 1000\)

\(= 903.1 \approx 903\)  (approx)