The weight of certain population of young females is approximately normally distributed with the mean of 66 kg and a standard deviation of 7.5 kg. If an individual is selected at random from this population, find the percentage of women who will weigh less than 64.5 kg.
To find the percentage of women who will weigh less than 64.5 kg, we need to calculate the z-score corresponding to 64.5 kg.
The z-score formula is: z = (x - μ) / σ
where:
x is the value we want to find the z-score for (64.5 kg),
μ is the mean of the population (66 kg),
σ is the standard deviation of the population (7.5 kg).
Substituting the given values, we have:
z = (64.5 - 66) / 7.5
Calculating this equation, we find:
z = -0.2
To find the percentage of women who weigh less than 64.5 kg, we need to find the cumulative probability associated with the z-score of -0.2. This can be done using a standard normal distribution table or a calculator.
Using a standard normal distribution table, the cumulative probability associated with a z-score of -0.2 is approximately 0.4207. This means that approximately 42.07% of women will weigh less than 64.5 kg.
To find the percentage of women who will weigh less than 64.5 kg, we need to calculate the z-score first.
The z-score formula is given by:
z = (x - μ) / σ
where:
x = the value we want to find the percentage for (64.5 kg)
μ = the mean of the distribution (66 kg)
σ = the standard deviation of the distribution (7.5 kg)
Plugging in the values, we get:
z = (64.5 - 66) / 7.5
Calculating this, we find:
z ≈ -0.2
Now, we need to find the percentage of data below this z-score using a standard normal distribution table or calculator.
Looking up the z-score of -0.2 in the standard normal distribution table or using a calculator, we find that the percentage of data below this z-score is approximately 42.17%.
Therefore, the percentage of women who will weigh less than 64.5 kg is approximately 42.17%.
To find the percentage of women who will weigh less than 64.5 kg, we can use the concept of standard normal distribution.
First, we need to convert the weight of 64.5 kg to a standard score, also known as a z-score. The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- x is the value we want to convert (in this case, 64.5 kg),
- μ is the mean of the population (in this case, 66 kg),
- σ is the standard deviation of the population (in this case, 7.5 kg).
Substituting the given values into the formula, we get:
z = (64.5 - 66) / 7.5
Now we can calculate the z-score:
z = -1.5 / 7.5
z = -0.2
Using a standard normal distribution table or a calculator, we can find the percentage of values that fall to the left of a z-score of -0.2.
Looking up the z-score of -0.2 in a standard normal distribution table, we get the value of 0.4207. This means that approximately 42.07% of the population will weigh less than 64.5 kg.
Therefore, the percentage of women who will weigh less than 64.5 kg is approximately 42.07%.