A normal distribution has a mean of 50 and a standard deviation of 4. Determine the value below which 95% of the observations will occur.
To determine the value below which 95% of the observations will occur in a normal distribution, we need to use the Z-score formula.
The Z-score formula is:
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the value we want to find
μ is the mean of the distribution
σ is the standard deviation of the distribution
In this case, we want to find the value below which 95% of the observations will occur. To do this, we need to find the Z-score corresponding to a cumulative probability of 0.95.
Using a standard normal distribution table or a calculator, we find that the Z-score corresponding to a cumulative probability of 0.95 is approximately 1.96.
Now we can use the Z-score formula to find the value below which 95% of the observations will occur.
1.96 = (X - 50) / 4
Rearranging the formula to solve for X:
X - 50 = 1.96 * 4
X - 50 = 7.84
X = 7.84 + 50
X = 57.84
Therefore, the value below which 95% of the observations will occur is approximately 57.84.
To determine the value below which 95% of the observations will occur in a normal distribution, we can use the concept of z-scores.
A z-score represents the number of standard deviations a particular value is from the mean of the distribution. By looking up the z-score associated with a cumulative probability of 0.95 (95%), we can find the corresponding value.
To determine the z-score, we can use the standard normal distribution table or a calculator that provides z-score calculations. However, to find the value below which 95% of the observations will occur, we need to find the z-score that corresponds to an area of 0.95 in the z-table, which corresponds to a cumulative probability of 0.5 + 0.45.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value we want to convert to a z-score
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
In this case, we want to find the z-score corresponding to a cumulative probability of 0.95 for a standard normal distribution with a mean (μ) of 0 and a standard deviation (σ) of 1.
To determine the value below which 95% of the observations will occur for a normal distribution with a mean of 50 and a standard deviation of 4, we need to find the z-score that corresponds to 0.95 when converted to a standard normal distribution.
Using the formula mentioned earlier, we can rearrange it to solve for x:
x = z * σ + μ
In this case:
- σ = 4 (standard deviation of the distribution)
- μ = 50 (mean of the distribution)
From the z-table, we find that the z-score for a cumulative probability of 0.95 is approximately 1.645.
Substituting the values into the equation, we have:
x = 1.645 * 4 + 50
x = 6.58 + 50
x = 56.58
Therefore, the value below which 95% of the observations will occur is approximately 56.58 in this normal distribution with a mean of 50 and a standard deviation of 4.