The medical records of infants delivered at Kaiser Memorial Hospital show that the infants' lengths at birth (in inches) are normally distributed with a mean of 22 and a standard deviation of 2.1. Find the probability that an infant selected at random from among those delivered at the hospital measures the following.
A)more than 24 in.
from your handy Z table, you find that
24 is 2/2.1 = 0.952 std above the mean
P(Z>0.952) = 0.17
See the app at
http://davidmlane.com/hyperstat/z_table.html
To find the probability that an infant selected at random from the hospital measures more than 24 inches at birth, we can use the standard normal distribution.
First, we need to convert the measurement of 24 inches into a standardized z-score using the formula:
z = (x - μ) / σ
Where:
x = the measurement we are interested in (24 inches)
μ = the mean of the distribution (22 inches)
σ = the standard deviation of the distribution (2.1 inches)
Plugging in the values into the formula:
z = (24 - 22) / 2.1
z = 2 / 2.1
z ≈ 0.9524
Next, we can use a z-table or a statistical software to find the probability associated with the z-score. The probability of an infant measuring more than 24 inches can be calculated as the complement of the cumulative probability up to the z-score:
P(X > 24) = 1 - P(X ≤ 24)
Using a standard normal distribution table or a statistical software,
P(X ≤ 24) ≈ 0.8292
Thus, the probability of an infant measuring more than 24 inches is:
P(X > 24) = 1 - 0.8292
P(X > 24) ≈ 0.1708
Therefore, the probability of an infant selected at random from among those delivered at the hospital measuring more than 24 inches is approximately 0.1708, or 17.08%.
To find the probability that an infant selected at random from among those delivered at the hospital measures more than 24 inches, we need to use the normal distribution.
1. To begin, we need to calculate the z-score (standard score) for 24 inches using the formula:
z = (x - μ) / σ
where x is the value we are interested in (24 inches in this case), μ is the mean (22 inches), and σ is the standard deviation (2.1 inches).
Plugging in the values, we get:
z = (24 - 22) / 2.1 = 0.95238
2. Once we have the z-score, we can use a standard normal distribution table or a statistical calculator to find the probability associated with that z-score.
Looking up 0.95238 in the standard normal distribution table, we find that the corresponding probability is 0.8289.
Note: If you're using a calculator, you can use the "normalcdf" function or similar functions to calculate the probability directly by specifying the lower bound (24) and the upper bound (infinity in this case).
3. However, we are interested in finding the probability of an infant measuring more than 24 inches, so we need to subtract the probability we found from 1.
P(X > 24) = 1 - 0.8289 = 0.1711
Therefore, the probability that an infant selected at random from among those delivered at the hospital measures more than 24 inches is approximately 0.1711 or 17.11%.