Question

a researcher measured the writing speed of 300 right-handed and 100 left-handed people. the right-handed group had a mean of 100 and a standard deviation of 8 while the left-handed group had a mean of 84 and a standard deviation of 10. assume the distributions are normal. how many left-handed people wrote faster than the average right-handed person did?

Explanation:

Step1: Calculate z-score for left-handers

The z-score formula is $z = \frac{X - \mu}{\sigma}$, where $X=100$ (avg right-handed speed), $\mu=84$ (left-handed mean), $\sigma=10$ (left-handed std dev).
$z = \frac{100 - 84}{10} = 1.6$

Step2: Find upper tail probability

Look up $z=1.6$ in standard normal table. The area to the left of $z=1.6$ is 0.9452. The area to the right (faster than 100) is $1 - 0.9452 = 0.0548$.

Step3: Calculate number of left-handers

Multiply probability by total left-handers (100).
$100 \times 0.0548 = 5.48$
Round to nearest whole number.

Answer:

Approximately 5 left-handed people wrote faster than the average right-handed person.