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 and 547.5 cm? Between what limits will the middle 50% of the observations lie?
Try using the normal distribution tool at http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html
You should get 2.3% of 1000 or 23 for the first question. The range is 2 or more standard deviations frm the mean.
Between 520.5 and 547.5, with is + or - 1 one standard deviation, you will get 68.3% of the samples or 683 (on average). Actual results will vary statistically.
The middle 50% are between 524 and 543
3/8 into a decimal
To answer these questions, we will use the properties of the normal distribution and the information provided.
1. How many observations are likely to exceed 561 cm?
We need to find the proportion of observations that exceed 561 cm. We can use the z-score formula to convert the 561 cm value into a z-score and then use a standard normal distribution table or calculator to find the proportion.
The z-score formula is given by:
z = (x - μ) / σ
where:
x = value of interest (561 cm in this case)
μ = mean (534 cm)
σ = standard deviation (13.5 cm)
Plugging in the values, we have:
z = (561 - 534) / 13.5
Calculating this, we get:
z ≈ 2
Using a standard normal distribution table or calculator, we can find the proportion of observations that have a z-score greater than 2. This corresponds to the area of the right tail of the distribution. Looking up the value for z = 2, we find that the proportion is approximately 0.0228.
So, approximately 0.0228 (or about 2.28%) of the observations are likely to exceed 561 cm.
2. How many will be between 520.5 cm and 547.5 cm?
To find the proportion of observations within a certain range, we need to convert both boundary values into z-scores and calculate the difference between the two z-scores.
For 520.5 cm:
z1 = (520.5 - 534) / 13.5
For 547.5 cm:
z2 = (547.5 - 534) / 13.5
Calculating these values, we get:
z1 ≈ -0.963
z2 ≈ 1
Now, we need to find the proportion of observations between z1 and z2. This corresponds to the area between the two z-scores in the standard normal distribution.
Using a standard normal distribution table or calculator, we can find these proportions:
Proportion between z1 and z2 = Proportion less than z2 - Proportion less than z1
Looking up the values for z1 and z2, we find:
Proportion less than z1 ≈ 0.167
Proportion less than z2 ≈ 0.841
So, the proportion of observations between 520.5 cm and 547.5 cm is approximately:
Proportion between z1 and z2 ≈ 0.841 - 0.167 ≈ 0.674
To find the actual number of observations within this range, we can multiply the proportion by the total number of observations (1000):
Number of observations ≈ 0.674 * 1000 ≈ 674
So, approximately 674 observations 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 refers to the range between the 25th percentile (the lower quartile or Q1) and the 75th percentile (the upper quartile or Q3). Since the normal distribution is symmetric, the mean (μ) will be the median, which is also the 50th percentile.
Using a standard normal distribution table or calculator, we can find the z-scores corresponding to the 25th and 75th percentiles.
For the 25th percentile:
z25 = -0.674
For the 75th percentile:
z75 = 0.674
Now, we can use these z-scores to find the corresponding values in the original distribution using the z-score formula:
Lower limit = μ + z25 * σ
Upper limit = μ + z75 * σ
Plugging in the values, we have:
Lower limit = 534 + (-0.674) * 13.5
Upper limit = 534 + 0.674 * 13.5
Calculating these values, we get:
Lower limit ≈ 524.099
Upper limit ≈ 543.901
So, the middle 50% of the observations will lie between approximately 524.099 cm and 543.901 cm.
To answer these questions, we need to use the properties of the normal distribution. The normal distribution is characterized by its mean and standard deviation.
Question 1: How many observations are likely to exceed 561 cm?
To find out how many observations are likely to exceed 561 cm, we need to calculate the z-score for this value and find the area under the curve to the right of the z-score. The z-score is calculated using the formula: z = (x - mean) / standard deviation.
Let's calculate the z-score for 561 cm:
z = (561 - 534) / 13.5 = 2
To find the area under the curve to the right of the z-score of 2, we can refer to the standard normal distribution table or use statistical software.
By looking up the z-score of 2 in the standard normal distribution table, we find the corresponding area is approximately 0.0228. This means that 2.28% of the observations are likely to exceed 561 cm.
To find the number of observations, we multiply the probability (0.0228) by the total number of observations (1000):
Number of observations = 0.0228 * 1000 = 22.8 (approximately)
Therefore, approximately 23 observations are likely to exceed 561 cm.
Question 2: How many will be between 520.5 cm and 547.5 cm?
Similarly, to find the number of observations between 520.5 cm and 547.5 cm, we need to calculate the z-scores for these values and find the area under the curve between these z-scores. Let's calculate the z-scores for 520.5 cm and 547.5 cm:
z1 = (520.5 - 534) / 13.5 = -0.981
z2 = (547.5 - 534) / 13.5 = 1
To find the area between these z-scores, we can subtract the cumulative probability for z1 from the cumulative probability for z2.
Using the standard normal distribution table or statistical software, we find that the cumulative probability for z1 is approximately 0.1645 and for z2 is approximately 0.8413.
Area between z1 and z2 = 0.8413 - 0.1645 = 0.6768
To find the number of observations, we multiply the probability (0.6768) by the total number of observations (1000):
Number of observations = 0.6768 * 1000 = 676.8 (approximately)
Therefore, approximately 677 observations will be between 520.5 cm and 547.5 cm.
Question 3: Between what limits will the middle 50% of the observations lie?
The middle 50% of the observations refers to the range that contains the central 50% of the data. This range is also known as the interquartile range (IQR). To find the limits of the middle 50% of the observations, we need to calculate the z-scores corresponding to the cumulative probabilities at the 25th and 75th percentiles.
The z-scores at the 25th and 75th percentiles divide the distribution into four equal parts, each containing 25% of the data. By referring to the standard normal distribution table or using statistical software, we find that the z-score at the 25th percentile is approximately -0.674 and at the 75th percentile is approximately 0.674.
To find the corresponding observation values, we can use the formula:
x = (z * standard deviation) + mean
Lower limit = (-0.674 * 13.5) + 534 = 524.141 cm (approximately)
Upper limit = (0.674 * 13.5) + 534 = 543.459 cm (approximately)
Therefore, the middle 50% of the observations will lie between approximately 524.141 cm and 543.459 cm.