Question

A computer company produces an extremely light laptop and they claim it is the lightest on the market weighing only 32 ounces. the actual weight of the laptop in the box has a normal distribution, with a mean of 33 oz. and a standard deviation of 0.7 oz. what proportion of laptops are overweight (i.e., weigh more than 32 oz.)?

Answer 1

Solution: We are given:

$\mu=0.33, \sigma =0.7$

Let $x$ be the weight (oz) of laptop

We have to find $P(x>32)$

To find the this probability, we need to find the z score value.

The z score is given below:

$z=\frac{x-\mu}{\sigma}$

       $=\frac{32-33}{0.7}$

       $=-1.43$

Now, we have to find $P(z>-1.43)$

Using the standard normal table, we have:

$P(z>-1.43)=0.9236$

0.9236 or 92.36% of laptops are overweight