.) A hospital in a large city records the weight of every infant born at the hospital. The distribution of weights is normally shaped, with a mea n µ= 2. 9 kilograms and a standard deviation o- = 0 .45. Determine the following:

a. The percentage of infants who weighed less than 2.1 kilograms
b. The percentile rank of a weight of 4.2 kilograms
c. The percentage of infants who weighed between 1.8 and 4.0 kilograms
d. The percentage of infants who weighed between 3.4 and 4.1 kilograms
e. The weight that divides the distribution such that 1% of the weights are above it
f. Beyond what weights do the most extreme 5% of the scores lie?
g. If 15,000 infants have been born at the hospital, how many weighed less than 3 .5 kilograms?

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Tutor

To solve these questions, we need to use the properties of the normal distribution. The normal distribution is a continuous probability distribution that is symmetric and bell-shaped.

a. To find the percentage of infants who weighed less than 2.1 kilograms, we need to calculate the z-score and look up the corresponding area under the standard normal distribution curve.

First, we calculate the z-score using the formula:
z = (x - µ) / σ
where x is the weight we want to find the percentage for, µ is the mean, and σ is the standard deviation.

In this case, x = 2.1, µ = 2.9, and σ = 0.45.

z = (2.1 - 2.9) / 0.45
z = -1.78

Next, we look up the area to the left of -1.78 on the standard normal distribution table. The percentage of infants weighing less than 2.1 kilograms is equal to this area.

Using the table, we find that the area to the left of -1.78 is approximately 0.0359.

Therefore, the percentage of infants who weighed less than 2.1 kilograms is 3.59%.

b. The percentile rank of a weight of 4.2 kilograms represents the percentage of weights that are less than or equal to 4.2 kilograms. To find this, we calculate the z-score for 4.2 kilograms and use the standard normal distribution table to find the corresponding area.

Using the same formula as above, we find the z-score:
z = (4.2 - 2.9) / 0.45
z = 2.89

Next, we look up the area to the left of 2.89 on the standard normal distribution table. The percentile rank represents the percentage of weights that are less than or equal to 4.2 kilograms.

Using the table, we find that the area to the left of 2.89 is approximately 0.9987.

Therefore, the percentile rank of a weight of 4.2 kilograms is 99.87%.

c. To find the percentage of infants who weighed between 1.8 and 4.0 kilograms, we need to calculate the areas to the left of 1.8 and 4.0 on the standard normal distribution curve, and then take the difference between these two areas.

First, we calculate the z-scores for 1.8 and 4.0 kilograms:
z1 = (1.8 - 2.9) / 0.45
z1 = -2.44
z2 = (4.0 - 2.9) / 0.45
z2 = 2.44

Next, we look up the areas to the left of -2.44 and 2.44 on the standard normal distribution table. The percentage of infants weighing between 1.8 and 4.0 kilograms is equal to the difference between these two areas.

Using the table, we find that the area to the left of -2.44 is approximately 0.0074 and the area to the left of 2.44 is approximately 0.9926.

Therefore, the percentage of infants who weighed between 1.8 and 4.0 kilograms is 99.26% - 0.74% = 98.52%.

d. To find the percentage of infants who weighed between 3.4 and 4.1 kilograms, we follow the same steps as in part c.

First, we calculate the z-scores for 3.4 and 4.1 kilograms:
z1 = (3.4 - 2.9) / 0.45
z1 = 1.11
z2 = (4.1 - 2.9) / 0.45
z2 = 2.67

Next, we look up the areas to the left of 1.11 and 2.67 on the standard normal distribution table. The percentage of infants weighing between 3.4 and 4.1 kilograms is equal to the difference between these two areas.

Using the table, we find that the area to the left of 1.11 is approximately 0.8665 and the area to the left of 2.67 is approximately 0.9965.

Therefore, the percentage of infants who weighed between 3.4 and 4.1 kilograms is 99.65% - 86.65% = 12%.

e. The weight that divides the distribution such that 1% of the weights are above it can be found by finding the z-score corresponding to the area of 0.99 (1 - 0.01) in the standard normal distribution table.

Using the table, we find that the z-score corresponding to an area of 0.99 is approximately 2.33.

To find the weight, we use the formula:
x = µ + (z * σ)
where x is the weight we want to find, µ is the mean, σ is the standard deviation, and z is the z-score.

x = 2.9 + (2.33 * 0.45)
x = 3.8365

Therefore, the weight that divides the distribution such that 1% of the weights are above it is approximately 3.8365 kilograms.

f. To find beyond what weights do the most extreme 5% of the scores lie, we need to determine the z-score corresponding to the area of 0.05/2 (since it is divided into the two tails of the distribution) in the standard normal distribution table.

Using the table, we find that the z-score corresponding to an area of 0.025 (0.05/2) is approximately -1.96 (for the left tail) and 1.96 (for the right tail).

To find the weights, we use the formula:
x = µ + (z * σ)
where x is the weight we want to find, µ is the mean, σ is the standard deviation, and z is the z-score.

For the left tail:
x1 = 2.9 + (-1.96 * 0.45)
x1 = 1.171

For the right tail:
x2 = 2.9 + (1.96 * 0.45)
x2 = 4.229

Therefore, the most extreme 5% of the scores lie beyond the weights of approximately 1.171 kilograms (left tail) and 4.229 kilograms (right tail).

g. To find how many infants weighed less than 3.5 kilograms out of 15,000 infants, we need to calculate the z-score for 3.5 kilograms, and then multiply it by the total number of infants.

First, we calculate the z-score for 3.5 kilograms using the formula:
z = (x - µ) / σ
where x is 3.5, µ is 2.9, and σ is 0.45.

z = (3.5 - 2.9) / 0.45
z = 1.33

Next, we multiply the z-score by the total number of infants:
Number of infants = z * Total number of infants
Number of infants = 1.33 * 15000
Number of infants = 19,950

Therefore, out of 15,000 infants, approximately 19,950 infants weighed less than 3.5 kilograms.

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z scores. Multiply by 100 to get percentages.

F. Reverse process to get Z score for .05, the use the equation above.

G. multiply proportion found by 15,000.