3. 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?

My favourite webpage for this type of question is

http://davidmlane.com/hyperstat/z_table.html

Just enter the data as is, you don't even need to convert them to z scores.

To find out how many observations are likely to exceed 561 cm, we can use the standard normal distribution table.

Step 1: Calculate the z-score for 561 cm using the formula:
z = (x - μ) / σ
Where x is the value, μ is the mean, and σ is the standard deviation.

z = (561 - 534) / 13.5
z ≈ 2

Step 2: Look up the corresponding area under the standard normal distribution curve for a z-score of 2. From the standard normal distribution table, the area to the right of z = 2 is approximately 0.0228.

Step 3: Multiply the area by the total number of observations to find the number of observations likely to exceed 561 cm:
Number of observations = 0.0228 * 1000 ≈ 22.8

Since the number of observations should be a whole number, we can expect around 23 observations to exceed 561 cm.

To find the number of observations between 520.5 cm and 547.5 cm, we can use the same approach.

Step 1: Calculate the z-scores for the lower and upper values:
Lower z = (520.5 - 534) / 13.5 ≈ -0.963
Upper z = (547.5 - 534) / 13.5 ≈ 1

Step 2: Look up the area under the standard normal distribution curve for these z-scores.
For the lower z-value (-0.963), the area is approximately 0.1671.
For the upper z-value (1), the area is approximately 0.8413.

Step 3: Subtract the lower area from the upper area to get the total area between the values:
Total area between = 0.8413 - 0.1671 ≈ 0.6742

Step 4: Multiply the area by the total number of observations to find the number of observations in this range:
Number of observations = 0.6742 * 1000 ≈ 674.2

Again, since the number of observations should be a whole number, we can expect around 674 observations to be between 520.5 cm and 547.5 cm.

Lastly, to find the limits within which the middle 50% of observations lie, we need to determine the z-scores for the corresponding percentiles.

Step 1: Find the z-scores for the lower and upper percentiles:
Lower percentile: (100 - 50) / 2 = 25
Upper percentile: 100 - ((100 - 50) / 2) = 75

Step 2: Look up the z-scores corresponding to these percentiles in the standard normal distribution table.
The z-score for the 25th percentile is approximately -0.6745, and the z-score for the 75th percentile is approximately 0.6745.

Step 3: Calculate the corresponding values using the z-scores:
Lower value = (z * σ) + μ = (-0.6745 * 13.5) + 534 ≈ 524.4 cm
Upper value = (z * σ) + μ = (0.6745 * 13.5) + 534 ≈ 543.6 cm

Therefore, the middle 50% of the observations will lie between 524.4 cm and 543.6 cm.

To solve this problem, we need to use the properties of the normal distribution and the given information about the mean and standard deviation.

1. How many observations are likely to exceed 561 cm?
To determine the number of observations likely to exceed a certain value, we need to calculate the area under the normal distribution curve for values greater than 561 cm.

We can use a standard Z-score formula: Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.

In this case, we have X = 561 cm, μ = 534 cm, and σ = 13.5 cm.

Calculating the Z-score: Z = (561 - 534) / 13.5 = 2

Now, we look up the Z-score in a standard normal distribution table (also known as a Z-table) to find the corresponding area.

The Z-table gives us the area to the left of the Z-score. Since we want the area to the right of the Z-score, we subtract the Z-score's area from 1.

In the Z-table, a Z-score of 2 corresponds to an area of 0.9772.

Therefore, the probability of an observation exceeding 561 cm is 1 - 0.9772 = 0.0228.

To find the number of observations likely to exceed 561 cm, we multiply this probability by the total number of observations, which in this case is 1000.

Number of observations likely to exceed 561 cm = 0.0228 * 1000 ≈ 22.8

Since the number of observations must be a whole number, we can conclude that approximately 23 observations are likely to be greater than 561 cm.

2. How many observations will be between 520.5 cm and 547.5 cm?
To find the number of observations within a specific range, we need to calculate the area under the normal distribution curve between the two values.

We will use the same Z-score formula and the given values:
For X1 = 520.5 cm: Z1 = (520.5 - 534) / 13.5 = -0.963
For X2 = 547.5 cm: Z2 = (547.5 - 534) / 13.5 = 1

Now, we find the areas corresponding to these Z-scores using the Z-table:
For Z1 = -0.963, the area is 0.1664
For Z2 = 1, the area is 0.8413

To calculate the number of observations between the two values, we need to find the difference between the two areas and multiply it by the total number of observations.

Number of observations between 520.5 cm and 547.5 cm = (0.8413 - 0.1664) * 1000 ≈ 674.9

Since the number of observations must be a whole number, we can conclude that approximately 675 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 corresponds to the interquartile range (IQR), which includes the range from the 25th to the 75th percentile. In a normal distribution, these percentiles are symmetric around the mean.

To find the Z-scores corresponding to these percentiles, we look up the Z-scores associated with the cumulative probabilities of 0.25 and 0.75 in the Z-table.

For the 25th percentile, we find Z1 = -0.6745, and for the 75th percentile, we find Z2 = 0.6745.

Now, we use the Z-score formula to calculate the corresponding X-values:
For X1: -0.6745 = (X1 - 534) / 13.5
X1 = -0.6745 * 13.5 + 534 ≈ 524.07

For X2: 0.6745 = (X2 - 534) / 13.5
X2 = 0.6745 * 13.5 + 534 ≈ 543.54

Therefore, the middle 50% of the observations will lie between approximately 524.07 cm and 543.54 cm.