Questions: Use the normal distribution of SAT critical reading scores for which the mean is 510 and the standard deviation is 110. Assume the variable x is normally distributed. (a) What percent of the SAT verbal scores are less than 625? (b) If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 550? (a) Approximately 85.21% of the SAT verbal scores are less than 625. (Round to two decimal places as needed.) (b) You would expect that approximately SAT verbal scores would be greater than 550. (Round to the nearest whole number as needed.)

Use the normal distribution of SAT critical reading scores for which the mean is 510 and the standard deviation is 110. Assume the variable x is normally distributed.

(a) What percent of the SAT verbal scores are less than 625?

(b) If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 550?

(a) Approximately 85.21% of the SAT verbal scores are less than 625. (Round to two decimal places as needed.)

(b) You would expect that approximately SAT verbal scores would be greater than 550. (Round to the nearest whole number as needed.)

Transcript text: Use the normal distribution of SAT critical reading scores for which the mean is 510 and the standard deviation is 110 . Assume the variable x is normally distributed. (a) What percent of the SAT verbal scores are less than 625 ? (b) If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 550 ? Click to view page 1 of the standard normal table. Click to view page 2 of the standard normal table. (a) Approximately $85.21 \%$ of the SAT verbal scores are less than 625. (Round to two decimal places as needed.) (b) You would expect that approximately $\square$ SAT verbal scores would be greater than 550. (Round to the nearest whole number as needed.)

Solution

Solution Steps

Step 1: Standardize the Specific Value

To standardize the specific value $x = 625$, we calculate the Z-score using the formula $Z = \frac{x - \mu}{\sigma}$, where $\mu = 510$ and $\sigma = 110$. Substituting the given values, we get $Z = \frac{625 - 510}{110} = 1.045$.

Step 2: Use the Standard Normal Distribution

Using the Z-score of 1.045, we refer to the standard normal distribution table to find the cumulative probability of 0.852. This means that 85.21% of observations are expected to fall below $x = 625$ and 14.79% above it.

Step 3: Calculate the Expected Number

With a sample size of $n = 1000$, we expect 852.09 observations to fall below $x = 625$ and 147.91 above it, rounded to 2 decimal places.

Final Answer:

Percentage below $x = 625$: 85.21% Expected number below $x = 625$: 852.09 Percentage above $x = 625$: 14.79% Expected number above $x = 625$: 147.91

Step 1: Standardize the Specific Value

To standardize the specific value $x = 550$, we calculate the Z-score using the formula $Z = \frac{x - \mu}{\sigma}$, where $\mu = 510$ and $\sigma = 110$. Substituting the given values, we get $Z = \frac{550 - 510}{110} = 0.364$.

Step 2: Use the Standard Normal Distribution

Using the Z-score of 0.364, we refer to the standard normal distribution table to find the cumulative probability of 0.642. This means that 64% of observations are expected to fall below $x = 550$ and 36% above it.

Step 3: Calculate the Expected Number

With a sample size of $n = 1000$, we expect 642 observations to fall below $x = 550$ and 358 above it, rounded to 0 decimal places.

Final Answer:

Percentage below $x = 550$: 64% Expected number below $x = 550$: 642 Percentage above $x = 550$: 36% Expected number above $x = 550$: 358

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