# Percentage of Packets given Mean and Standard Deviation

The Weight of a packet of imported biscuits from a shipment is normally distributed with a mean of 500g and a standard deviation of 40g. What percentage of packets of the shipment weighs between 540good and 560g?

• Thank you Bruce....am doing my best...first time working on these and will put more time. Much appreciated – Thamsanqa Mlungwana Apr 17 '17 at 7:14
• Sorry. Fixing typo in earlier comment. Start with $P(540 < X < 560) = P(\frac{540-500}{40} < Z <\frac{560-500}{40}),$ where $Z \sim \mathsf{Norm}(0,1).$ Then use printed table of standard normal CDF. By software I get about 9%. For example, can you use the printed table to find $P(0 < Z < 1)$ and $P(0 < Z < 1.5)?$ – BruceET Apr 17 '17 at 7:29
• Make a sketch. Past bedtime here. Will check in the morning. – BruceET Apr 17 '17 at 7:36

$$P(540 < X < 560) = P\left(\frac{540-500}{40} < \frac{X - \mu}{\sigma} < \frac{560-500}{40}\right) = P(1 < Z < 1.5)\\ = P(0 < Z < 1.5) - P(0 < Z < 1) = 0.9332 - 0.8413 = 0.0919,$$ where you should be able to get $P(0 < Z < 1.5)$ and $P(0 < Z < 1)$ from printed standard normal tables.