To find the mean ($\mu$) and standard deviation ($\sigma$) of the normal distribution, we can use the information given:

1. 1.25% of the area lies to the left of 42: This implies that 42 is at the 1.25th percentile of the distribution.
2. 1.25% of the area lies to the right of 48: This implies that 48 is at the 98.75th percentile of the distribution.

Steps to Solve

1. Determine the Z-scores: Since 1.25% lies in the left tail and 1.25% in the right tail, we can find the Z-scores corresponding to these percentiles in a standard normal distribution.

   • For the 1.25th percentile (left tail), the Z-score is approximately $Z = -2.24$.
   • For the 98.75th percentile (right tail), the Z-score is approximately $Z = +2.24$.

2. Set up equations using the Z-score formula: The Z-score formula is: $Z = \frac{X - \mu}{\sigma}$ where:

   • $X$ is the value in the distribution,
   • $\mu$ is the mean,
   • $\sigma$ is the standard deviation.

   Since we have two values, 42 and 48, we set up two equations:

   • For $X = 42$ (1.25th percentile): $-2.24 = \frac{42 - \mu}{\sigma}$

   • For $X = 48$ (98.75th percentile): $2.24 = \frac{48 - \mu}{\sigma}$

3. Solve for $\mu$ and $\sigma$: We have a system of equations: $-2.24 \sigma = 42 - \mu$ $2.24 \sigma = 48 - \mu$

   Adding these two equations to eliminate $\mu$: $-2.24 \sigma + 2.24 \sigma = 42 + 48 - 2\mu$ Simpl