from a set of 1000 observations known to be normally distributed the mean is 534 cm and sd is 13.5 cm. how many observations are likely to exceed 561 cm, how many will be between 520.5 cm 547.5 cm, between what limits will the middle 50% of the observations lie
To find out how many observations are likely to exceed 561 cm, we need to calculate the z-score for 561 cm and then use the standard normal distribution table to find the probability.
Step 1: Calculate the z-score:
z = (x - μ) / σ
where x = 561 cm, μ = 534 cm, and σ = 13.5 cm.
z = (561 - 534) / 13.5
z = 2
Step 2: Use the standard normal distribution table to find the probability corresponding to a z-score of 2. The probability of a z-score greater than 2 is approximately 0.0228.
Step 3: Multiply the probability by the total number of observations to find the number of observations likely to exceed 561 cm:
Number of observations = Probability * Total number of observations
Number of observations = 0.0228 * 1000
Number of observations ≈ 22.8 (rounded to the nearest whole number)
Therefore, it is likely that approximately 23 observations will exceed 561 cm.
To find out how many observations will be between 520.5 cm and 547.5 cm, we again need to calculate the z-scores for these values and use the standard normal distribution table to find the probabilities.
Step 1: Calculate the z-scores:
For 520.5 cm:
z1 = (520.5 - 534) / 13.5
z1 ≈ -1
For 547.5 cm:
z2 = (547.5 - 534) / 13.5
z2 ≈ 1
Step 2: Use the standard normal distribution table to find the probabilities corresponding to these z-scores:
The probability for a z-score of -1 is approximately 0.1587.
The probability for a z-score of 1 is approximately 0.8413.
Step 3: Multiply the probabilities by the total number of observations to find the number of observations between 520.5 cm and 547.5 cm:
Number of observations = (Probability for z2) - (Probability for z1) * Total number of observations
Number of observations ≈ (0.8413 - 0.1587) * 1000
Number of observations ≈ 682
Therefore, there will be approximately 682 observations between 520.5 cm and 547.5 cm.
To find the limits within which the middle 50% of the observations lie, we need to find the z-scores that correspond to the lower and upper quartiles using the standard normal distribution table.
Step 1: Find the z-score for the lower quartile (25th percentile):
The lower quartile corresponds to a cumulative probability of 0.25.
Using the standard normal distribution table, the z-score for a cumulative probability of 0.25 is approximately -0.6745.
Step 2: Find the z-score for the upper quartile (75th percentile):
The upper quartile corresponds to a cumulative probability of 0.75.
Using the standard normal distribution table, the z-score for a cumulative probability of 0.75 is approximately 0.6745.
Step 3: Calculate the corresponding values for these z-scores:
For the lower quartile:
Value1 = μ + (z-score * σ)
Value1 = 534 + (-0.6745 * 13.5)
Value1 ≈ 524.53 (rounded to two decimal places)
For the upper quartile:
Value2 = μ + (z-score * σ)
Value2 = 534 + (0.6745 * 13.5)
Value2 ≈ 543.47 (rounded to two decimal places)
Therefore, the middle 50% of the observations will lie between approximately 524.53 cm and 543.47 cm.
To answer these questions, we can use the properties of the normal distribution and the given mean and standard deviation.
1. How many observations are likely to exceed 561 cm?
To find this, we need to calculate the area under the normal curve to the right of 561 cm.
We can use a z-score to standardize the value and then consult a standard normal distribution table or use a calculator to find the corresponding area. The z-score is calculated as:
z = (x - μ) / σ,
where x is the value we are interested in, μ is the mean, and σ is the standard deviation.
In this case,
z = (561 - 534) / 13.5 = 2.
Using a standard normal distribution table or calculator, we can find that the area to the right of z = 2 is approximately 0.0228 or 2.28%.
Therefore, we can estimate that about 2.28% of the observations (1000 * 0.0228) are likely to exceed 561 cm.
2. How many observations will be between 520.5 cm and 547.5 cm?
To find this, we need to calculate the area under the normal curve between these two values.
First, we calculate the z-scores for both limits:
z1 = (520.5 - 534) / 13.5 = -0.964,
z2 = (547.5 - 534) / 13.5 = 1.
Again, using a standard normal distribution table or calculator, we find the areas corresponding to these z-scores:
Area to the left of z1 ≈ 0.1664 or 16.64%,
Area to the left of z2 ≈ 0.8413 or 84.13%.
To find the area between these two z-scores, we subtract the lower area from the higher area:
Area = 0.8413 - 0.1664 ≈ 0.6749 or 67.49%.
Therefore, we can estimate that about 67.49% of the observations (1000 * 0.6749) will be between 520.5 cm and 547.5 cm.
3. Between what limits will the middle 50% of the observations lie?
The middle 50% of the observations is also known as the interquartile range. To find the limits for this range, we need to consider the z-scores.
The z-score corresponding to the lower limit is determined by finding the area below it that represents 25%:
Area = 0.25,
z = the z-score corresponding to 25% area.
Similarly, the z-score corresponding to the upper limit is determined by finding the area below it that represents 75%:
Area = 0.75,
z = the z-score corresponding to 75% area.
Using a standard normal distribution table or calculator, we can find that the z-score corresponding to a 25% area is approximately -0.6745, and the z-score corresponding to a 75% area is approximately 0.6745.
Now, we can calculate the actual values corresponding to these z-scores:
x1 = μ + z1 * σ = 534 + (-0.6745) * 13.5 ≈ 523.10 cm,
x2 = μ + z2 * σ = 534 + 0.6745 * 13.5 ≈ 544.90 cm.
Therefore, the middle 50% of the 1000 observations is likely to lie between approximately 523.10 cm and 544.90 cm.