The average age of a child in kindergarden is 5 years or 60 months. The standard deviation is 12 months. If a random child is selected, find the probability that the childs age is between 54 and 66 months.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores. You will want to add the proportions between the two Z scores and the mean.
To find the probability that a randomly selected child's age is between 54 and 66 months, we need to use the concept of the standard normal distribution.
Step 1: Calculate the z-scores for the given ages.
The z-score calculates how many standard deviations an individual value is from the mean. It can be calculated using the formula:
z = (x - μ) / σ
Where:
x = the age value
μ = the mean age (in months)
σ = the standard deviation (in months)
For 54 months:
z₁ = (54 - 60) / 12
z₁ = -0.5
For 66 months:
z₂ = (66 - 60) / 12
z₂ = 0.5
Step 2: Look up the z-scores in the standard normal distribution table.
The standard normal distribution table provides the probability associated with each z-score.
From the table, the probability associated with a z-score of -0.5 is approximately 0.3085, and the probability associated with a z-score of 0.5 is also approximately 0.3085.
Step 3: Calculate the probability between the two z-scores.
Since the z-scores represent the probability of each individual value, we need to calculate the probability between the two z-scores.
P(54 < x < 66) = P(x < 66) - P(x < 54)
= 0.3085 - 0.3085
= 0
Therefore, the probability that a randomly selected child's age is between 54 and 66 months is 0.