The mean score on a normally distributed exam is 42 with a standard deviation of 12.1. Which socre would be expected to occur less than 5% of the time?
1) 25
2) 32
3) 60
4) 67
-I know the answer is 67, but i don't know why .. PLEASE help!
is 67 outside two standard deviations?
Im not sure , but say if it was... can u explain it to me?
To determine which score would be expected to occur less than 5% of the time, we need to use the concept of z-scores.
A z-score measures the number of standard deviations a given value is from the mean. We can use the formula:
z = (x - μ) / σ
Where:
x = the value you want to find the probability for
μ = the mean of the distribution
σ = the standard deviation of the distribution
First, we need to calculate the z-score for each option using the given values:
For option 1:
z1 = (25 - 42) / 12.1
z1 ≈ -1.4
For option 2:
z2 = (32 - 42) / 12.1
z2 ≈ -0.83
For option 3:
z3 = (60 - 42) / 12.1
z3 ≈ 1.49
For option 4:
z4 = (67 - 42) / 12.1
z4 ≈ 2.07
Next, we need to find the corresponding probability for each z-score using a z-table (also known as a standard normal distribution table). This table gives the area under the normal distribution curve to the left of a given z-score.
Looking up the z-scores in the table, we can find the probabilities:
For option 1:
P(z < -1.4) ≈ 0.0808
For option 2:
P(z < -0.83) ≈ 0.2033
For option 3:
P(z < 1.49) ≈ 0.9281
For option 4:
P(z < 2.07) ≈ 0.9812
Now, we can see that the option with a probability less than 5% is option 4, where P(z < 2.07) ≈ 0.9812. Therefore, the score of 67 would be expected to occur less than 5% of the time.
To determine which score would be expected to occur less than 5% of the time, we can use the concept of z-scores and the standard normal distribution.
Step 1: Calculate the z-score
The z-score represents the number of standard deviations a particular data point is away from the mean. It is calculated using the formula:
z = (x - μ) / σ
where:
- x is the data point
- μ is the mean
- σ is the standard deviation
In this case, the mean (μ) is 42, and the standard deviation (σ) is 12.1.
For each of the given scores, we need to calculate the corresponding z-score.
For option 1) 25:
z = (25 - 42) / 12.1 = -1.404
For option 2) 32:
z = (32 - 42) / 12.1 = -0.826
For option 3) 60:
z = (60 - 42) / 12.1 = 1.487
For option 4) 67:
z = (67 - 42) / 12.1 = 2.066
Step 2: Find the corresponding area under the standard normal curve
The next step is to find the area (probability) associated with each z-score. This area represents the proportion of scores that are less than a particular z-score.
We can refer to a standard normal distribution table (also known as a z-table) or use statistical software to find the area.
For option 1) 25: The z-score is -1.404. Looking up this value on the z-table, you would find an area of approximately 0.0811 or 8.11%.
For option 2) 32: The z-score is -0.826. The corresponding area from the z-table is approximately 0.2033 or 20.33%.
For option 3) 60: The z-score is 1.487. The approximate area from the z-table is 0.9309 or 93.09%.
For option 4) 67: The z-score is 2.066. Looking up this value on the z-table, you would find an area of approximately 0.9807 or 98.07%.
Step 3: Determine the score that occurs less than 5% of the time
Since we are looking for a score that occurs less than 5% of the time, we need to find the option with an area (probability) less than 0.05.
From the calculated areas above, we can see that option 4) 67 has an area of approximately 0.9807 or 98.07%. Since this area is greater than 0.05 (5%), it means that only 98.07% of scores are expected to be less than 67.
Hence, the correct answer is option 4) 67 - expected to occur less than 5% of the time.