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 answer these questions, we will use the properties of the normal distribution. The normal distribution is a continuous probability distribution that is symmetrical and typically represented by a bell-shaped curve.

Question 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 find the area under the normal curve beyond that value. In this case, we want to find the area to the right of 561 cm.

First, we need to standardize the value of 561 cm to a Z-score using the formula:

Z = (X - μ) / σ

Where:
X = the value we want to standardize (561 cm)
μ = the mean (534 cm)
σ = the standard deviation (13.5 cm)

Plugging in the values, we get:

Z = (561 - 534) / 13.5

Z ≈ 2

Now we can use a standard normal distribution table or calculator to find the area to the right of Z = 2. In most tables, you will find this as P(Z > 2). The corresponding value is approximately 0.0228.

This means that approximately 2.28% of the observations are likely to exceed 561 cm in a set of 1000 normally distributed observations.

Question 2: How many observations will be between 520.5 cm and 547.5 cm?

To find the number of observations between two values, we need to find the area under the normal curve between these values. We can calculate this by standardizing the values to Z-scores and finding the difference between their cumulative probabilities.

First, we standardize the values:

Z_lower = (520.5 - 534) / 13.5
Z_upper = (547.5 - 534) / 13.5

Z_lower ≈ -0.963
Z_upper ≈ 0.963

Next, we use the standard normal distribution table or calculator to find the cumulative probability for each Z-score. Let's denote this probability as P(Z).

P_lower = P(Z < -0.963)
P_upper = P(Z < 0.963)

By subtracting the two probabilities, we get the desired area:

P_between = P_upper - P_lower

Using the standard normal distribution table or calculator, we find:

P_between ≈ 0.8264

Therefore, approximately 82.64% of the observations will be between 520.5 cm and 547.5 cm in a set of 1000 normally distributed observations.

Question 3: Between what limits will the middle 50% of the observations lie?

The middle 50% of the observations correspond to the interquartile range (IQR). To find the limits, we need to determine the Z-scores that correspond to the 25th and 75th percentiles.

The 25th percentile (Q1) corresponds to a Z-score of -0.674.

The 75th percentile (Q3) corresponds to a Z-score of 0.674.

Now we need to convert these Z-scores back to their original values using the formula:

X = (Z * σ) + μ

X_lower = (-0.674 * 13.5) + 534
X_upper = (0.674 * 13.5) + 534

X_lower ≈ 524.05
X_upper ≈ 543.95

Therefore, the middle 50% of the observations will lie between approximately 524.05 cm and 543.95 cm in a set of 1000 normally distributed observations.