the scores of the students on a standardized test are normally distributed, with a mean of 500 and a…

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set. For $x = 350$, $\mu=500$, $\sigma = 110$, then $z_1=\frac{350 - 500}{110}=\frac{- 150}{110}\approx - 1.36$. For $x = 550$, $\mu = 500$, $\sigma=110$, then $z_2=\frac{550 - 500}{110}=\frac{50}{110}\approx0.45$.

Step2: Use the standard - normal table

The standard - normal table gives the cumulative probability $P(Z\leq z)$. We know that $P(-1.36<Z<0.45)=P(Z < 0.45)-P(Z < - 1.36)$. Since the standard - normal distribution is symmetric about $z = 0$, $P(Z < - 1.36)=1 - P(Z < 1.36)$. From the table, $P(Z < 0.45)=0.6736$ and $P(Z < 1.36)=0.9131$, so $P(Z < - 1.36)=1 - 0.9131 = 0.0869$. Then $P(-1.36<Z<0.45)=0.6736-0.0869 = 0.5867\approx59%$.

Answer:

59%