the wieights of newborn babies are distributed normally, with the mean of approximately 115 oz and a standard deviation of 20 oz. if a newborn baby is selected at random. what is the probability that the baby weighs more than 95 oz?
To find the probability that a randomly selected newborn baby weighs more than 95 oz, we need to use the normal distribution.
In this case, we are given the mean (μ) of the weight distribution, which is approximately 115 oz, and the standard deviation (σ), which is 20 oz. We can assume that the weights of newborn babies follow a normal distribution curve.
To find the probability that a baby weighs more than 95 oz, we need to calculate the area under the normal curve to the right of 95 oz. In other words, we need to find the area to the right of 95 oz on the normal distribution curve.
To do this, we can use the Z-score formula:
Z = (X - μ) / σ
Where:
X = the value we want to find the probability for (95 oz in this case)
μ = the mean of the distribution (115 oz)
σ = the standard deviation (20 oz)
Substituting the values, we have:
Z = (95 - 115) / 20
Z = -20 / 20
Z = -1
To find the probability associated with the Z-score of -1, we can refer to the Z-table or use a statistical software/tool.
By looking up the Z-score of -1 in the Z-table, we can find the corresponding probability. The Z-table gives us the area under the standard normal curve (Z-distribution), which has a mean of 0 and a standard deviation of 1.
The Z-table tells us that the probability associated with a Z-score of -1 is approximately 0.1587. However, since we are interested in the area to the right of 95 oz, we subtract this value from 1 to obtain the probability that the baby weighs more than 95 oz.
So, the probability that a randomly selected newborn baby weighs more than 95 oz is approximately 1 - 0.1587 = 0.8413, or 84.13%.
Note that in some statistical software/tools, you can directly input the values of X, μ, and σ to find the probability without having to look up the Z-table.