Answer :
The ages of people who own their homes are normally distributed with a mean of 42 and a standard deviation of 3.2. To find the age at which 75% of the home owners are older than this age, we can use the standard normal distribution table to find the z-score, which is approximately 0.674. The correct option is 44.2, as 75% of the home owners are older than 44.2.
Suppose ages of people who own their homes are normally distributed with a mean of 42 years and a standard deviation of 3.2 years. We need to find the age at which 75% of the home owners are older than this age.How to find the age?We can use the standard normal distribution table. The standard normal distribution is a normal distribution with a mean of 0 and standard deviation of 1. Using this table, we can find the z-score which corresponds to the area to the left of a particular value.Suppose z is the z-score such that the area to the left of z is 0.75. This means that 75% of the distribution is to the left of z. We can use this z-score to find the age at which 75% of the home owners are older than this age.What is the formula for z-score?The formula for finding the z-score for a value x in a normal distribution with mean μ and standard deviation σ is as follows
[tex]:$$z=\frac{x-\mu}{\sigma}$$[/tex]
We can rearrange this formula to find the value of x. Here's how:$$x=\sigma z + \mu$$
Now, let's find the z-score which corresponds to the area to the left of 0.75. Using a standard normal distribution table
we can find that this z-score is approximately 0.674.Using the above formula, we can find the age x at which 75% of the home owners are older than this age. We know that μ = 42 and σ = 3.2. Therefore, we have:$$x = 3.2 \times 0.674 + 42 = 44.16$$
Therefore, approximately 75% of the home owners are older than 44.16. So, the correct option is 44.2.Approximately 75% of the home owners are older than 44.2.
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