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 can use the properties of the normal distribution and the provided mean and standard deviation.

1. How many observations are likely to exceed 561 cm?
To find the number of observations likely to exceed a certain value in a normally distributed dataset, we need to calculate the z-score for that value and use the z-score table to find the corresponding cumulative probability. The formula for calculating the z-score is: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.

Let's calculate the z-score for 561 cm:
z = (561 - 534) / 13.5 ≈ 1.999

Looking up the z-score of 1.999 in a standard normal distribution table, we find that the cumulative probability associated with it is approximately 0.977. This means that approximately 97.7% of the observations are likely to fall below 561 cm. Therefore, the number of observations likely to exceed 561 cm can be estimated as (1 - 0.977) * 1000 ≈ 22 (rounding up to the nearest whole number).

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 z-scores for both values and use the z-score table to find the cumulative probabilities. Then, we subtract the lower cumulative probability from the higher cumulative probability to find the probability between the two values.

Let's calculate the z-scores for 520.5 cm and 547.5 cm:
z1 = (520.5 - 534) / 13.5 ≈ -1
z2 = (547.5 - 534) / 13.5 ≈ 1

Looking up the z-scores of -1 and 1 in the standard normal distribution table, we find the cumulative probabilities of approximately 0.1587 and 0.8413, respectively. Subtracting the lower cumulative probability from the higher, we get 0.8413 - 0.1587 ≈ 0.6826. This means that approximately 68.26% of the observations are likely to fall between 520.5 cm and 547.5 cm. Therefore, the number of observations in this range can be estimated as 0.6826 * 1000 ≈ 683 (rounding to the nearest whole number).

3. Between what limits will the middle 50% of the observations lie?
The middle 50% of the observations corresponds to the range between the 25th and 75th percentiles. To find these limits, we need to calculate the z-scores associated with those percentiles, and then use them to calculate the corresponding values in the dataset.

The 25th percentile corresponds to a cumulative probability of 0.25. To find the z-score for this probability, we can use the inverse of the cumulative distribution function (CDF) of the standard normal distribution. Using a z-score table, we find the z-score associated with a cumulative probability of 0.25 is approximately -0.6745.

Similarly, the 75th percentile corresponds to a cumulative probability of 0.75. The z-score for this probability is approximately 0.6745.

Now, we can calculate the corresponding values in the dataset:
Value1 = (z1 * σ) + μ = (-0.6745 * 13.5) + 534 ≈ 524.759 cm
Value2 = (z2 * σ) + μ = (0.6745 * 13.5) + 534 ≈ 543.241 cm

Therefore, the middle 50% of the observations will lie between approximately 524.759 cm and 543.241 cm.

To answer these questions, we need to use the properties of the standard normal distribution and z-scores. Follow these steps to calculate the desired values:

1. Calculate the z-score associated with each given observation using the formula:

z = (x - μ) / σ

where x is the given observation, μ is the mean, and σ is the standard deviation.

2. Use the z-score table or a calculator to find the area (probability) associated with each z-score. This represents the proportion of observations falling within a certain range.

Now let's solve each question step by step:

1. How many observations are likely to exceed 561 cm?

Find the z-score for 561 cm:

z = (561 - 534) / 13.5 = 2

The area to the right of z = 2 on the standard normal distribution curve is 0.0228 (from the z-table). This represents the proportion of observations exceeding 561 cm.

To find the number of observations, multiply the proportion by the total number of observations:

Number of observations = 0.0228 * 1000 = 22.8

Thus, approximately 23 observations can be expected to exceed 561 cm.

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

Find the z-scores for the given values:

z1 = (520.5 - 534) / 13.5 = -0.933
z2 = (547.5 - 534) / 13.5 = 1

The area between z1 and z2 represents the proportion of observations falling between 520.5 cm and 547.5 cm. Using the z-table, we find that the area (probability) from z1 to z2 is approximately 0.6915.

To find the number of observations, multiply the proportion by the total number of observations:

Number of observations = 0.6915 * 1000 = 691.5

Since we cannot have a fraction of an observation, we round down to the nearest whole number. Thus, approximately 691 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 lie within the two z-scores that enclose 25% on each side of the distribution.

To find the z-scores corresponding to the boundaries of the middle 50%, we look up the z-values associated with 0.25 and 0.75 in the z-table.

z1 = -z2 = ±z

The z-score corresponding to 0.25 is -0.674 and the z-score corresponding to 0.75 is 0.674.

Now we can calculate the actual values:

x1 = μ + z1 * σ
x2 = μ + z2 * σ

x1 = 534 + (-0.674) * 13.5 ≈ 524.46
x2 = 534 + 0.674 * 13.5 ≈ 543.54

The middle 50% of the observations are likely to fall between approximately 524.46 cm and 543.54 cm.