3. In a sample of 64 students given an examination in statistics, it is found that the mean weight is 57 kg with standard deviation 9 kg. Find :
Ć i) The proportion of students weight 48-62 kg.
Ć ii) The proportion of those less than 72 kg.
Ć iii) The proportion of those less than 50 kg ?
Ć iv) If the highest 5% of are over-weight, what is the Minimum score for an over-weight.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability for the Z scores.
For last problem, find Z score for 5% in smaller area and use its Z score.
To find the proportions in this problem, we will use the z-score formula and the standard normal distribution table.
The z-score formula is given by:
z = (x - μ) / σ
Where:
- z is the z-score,
- x is the value in question,
- μ is the mean, and
- σ is the standard deviation.
Now, let's calculate each proportion step by step:
i) The proportion of students weighing between 48 kg and 62 kg:
To find this proportion, we need to calculate the z-scores for 48 kg and 62 kg, and then find the area between these two z-scores in the standard normal distribution table.
First, let's calculate the z-score for 48 kg:
z1 = (48 - 57) / 9 = -1
Next, let's calculate the z-score for 62 kg:
z2 = (62 - 57) / 9 = 0.56
Looking up the values in the standard normal distribution table, we find that the area to the left of -1 is 0.1587, and the area to the left of 0.56 is 0.7123. To find the proportion between 48 kg and 62 kg, we subtract the smaller area from the larger area:
Proportion = 0.7123 - 0.1587 = 0.5536, or approximately 55.36%.
Therefore, the proportion of students weighing between 48 kg and 62 kg is approximately 55.36%.
ii) The proportion of students weighing less than 72 kg:
To find this proportion, we need to calculate the z-score for 72 kg and then find the area to the left of this z-score in the standard normal distribution table.
Let's calculate the z-score for 72 kg:
z = (72 - 57) / 9 = 1.67
Looking up the value in the standard normal distribution table, we find that the area to the left of 1.67 is 0.9525.
Therefore, the proportion of students weighing less than 72 kg is approximately 95.25%.
iii) The proportion of students weighing less than 50 kg:
To find this proportion, we need to calculate the z-score for 50 kg and then find the area to the left of this z-score in the standard normal distribution table.
Let's calculate the z-score for 50 kg:
z = (50 - 57) / 9 = -0.78
Looking up the value in the standard normal distribution table, we find that the area to the left of -0.78 is 0.2180.
Therefore, the proportion of students weighing less than 50 kg is approximately 21.80%.
iv) If the highest 5% of students are overweight, we need to find the z-score corresponding to the 95th percentile. This will give us the minimum score for being overweight.
Looking up the value in the standard normal distribution table, we find that the z-score corresponding to the 95th percentile is approximately 1.65.
To find the minimum score for being overweight, we use the z-score formula:
1.65 = (x - 57) / 9
Rearranging the formula, we solve for x:
x = (1.65 * 9) + 57
x = 71.85
Therefore, the minimum score for being overweight is approximately 71.85 kg.