Answer :
Answer:
Step-by-step explanation:
We can use the normal distribution table (also called the Z-table) to find the percentage of students who scored less than 34% in this scenario. Here's how:
Standardize the score:
We need to calculate the Z-score, which represents how many standard deviations a particular score (34%) is away from the mean (42%).
Z-score = (X - μ) / σ
where:
X - Individual score (34%)
μ - Mean (42%)
σ - Standard deviation (8%)
Z-score = (34% - 42%) / 8% = -1.00 (round to two decimal places)
Find the percentage using the Z-table:
Look up the Z-score -1.00 in a standard normal distribution table. The table will provide the area (probability) less than that Z-score.
Note: Most Z-tables only show positive Z-scores and the corresponding area. Since the normal distribution is symmetrical, the area less than a negative Z-score is equal to the area greater than its positive counterpart. In this case, the area less than -1.00 is the same as the area greater than 1.00.
Interpret the table value:
Look up 1.00 (or the closest value in the table) in the "greater than" section of the Z-table. This value represents the portion of the population that scored higher than 34%.
For example, if the table value for 1.00 is 0.8413, this means 84.13% of the students scored higher than 34%.
Calculate the percentage scoring less than 34%:
Since the normal distribution is symmetrical, the area less than the mean (42%) is equal to the area greater than the mean. Therefore, to find the percentage scoring less than 34%, subtract the area greater than 34% (obtained from the Z-table) from 1 (which represents the total area under the curve).
Percentage less than 34% = 1 - Area greater than 34% (from Z-table)
For example, if the area greater than 34% was 0.8413 (from step 3), then:
Percentage less than 34% = 1 - 0.8413 = 0.1587
Convert the decimal to percentage (optional):
Multiply the decimal value by 100% to express it as a percentage.
Percentage less than 34% ≈ 15.87% (round to two decimal places)
In conclusion, assuming a normal distribution, approximately 15.87% of the students scored less than 34% in the exam.
