Janine and Laura both took a standardized math test. The test scores were normally distributed with a mean of 75 and a standard deviation of 11. Janine earned a score of 76, whereas Laura earned a score of 89.
About what percentage of people had a score between Janine's and Laura's scores?
a) 30%
b) 43%
c) 80%
d) 36%
use a z-score table to find the population percentages
d) is probably good
To solve this problem, we need to calculate the z-scores for Janine's and Laura's scores and then use a standard normal distribution table to find the probability of having a score between their scores.
First, we need to calculate the z-scores. The formula is:
z = (x - μ) / σ
Where:
x = the score
μ = the mean
σ = the standard deviation
For Janine's score of 76:
z = (76 - 75) / 11 = 0.0909
For Laura's score of 89:
z = (89 - 75) / 11 = 1.2727
Next, we can use a standard normal distribution table (also known as the z-table) to find the probabilities associated with these z-scores.
Looking up the z-score of 0.0909 in the z-table, we find that it corresponds to a probability of 0.4641.
Looking up the z-score of 1.2727 in the z-table, we find that it corresponds to a probability of 0.8980.
To find the probability of having a score between Janine's and Laura's scores, we subtract the probability associated with Janine's z-score from the probability associated with Laura's z-score.
Probability = 0.8980 - 0.4641 = 0.4339, which is approximately 43%.
So, the answer is b) 43%.