The age of a group of 50 women are approximately normally distributed with a mea

Question

The age of a group of 50 women are approximately normally distributed with a mean of 48 years and a standard deviation of 5 years. One woman is randomly selected from the group and her age is observed.

a.       Find the probability that her age will fall between 55 and 60 years.

b.      Find the probability that her age will fall between 47 and 53 years.

c.       Find the probability that her age will be less than 35 years.

d.      Find the probability that her age will exceed 40 years.

View a table of areas under the standardized normal curve.

Explanation / Answer

It is doubtful in real life that you would take 50 ages and get exactly 51 for the average or exactly 5 for the standard deviation. I think they are trying to make the problem seem simpler, but something important is lost in the meaning.

A person's age is more accurately measured not just by full years but by years and days (possibly even smaller numbers) and the more exact the ages involved, the more sense it makes to use the normal distribution.

For example if a person happens to be 56 years 108 days and 6 hours old that would be x = 56 + 1/365 (108 + 6/24) = 56.297

Let's say it's a coincidence this time, and we'll use two decimal places to show the variables are continuous in the normal distribution.

So here are the steps:

There are two steps and the first step is to find the "z score" or the directed number of standard deviations away from the mean. Then the second step is to find the probability from a lookup table or calculator based on that "z score". I guess there is a third step to figure out what to do with the probabilities because the result of the lookup gives you the probability that the age is less than (whatever age you started with).

To clarify, how is to set it up and operate the Machine for Normal Distribution:

x: a fixed definite age at the end of the range (like 55.00 or 6.00 or 48.00 or 52.00 or 34.00 or 40.00)

?: the mean of the ages in this normal distribution= 51.00

?: the standard deviation of the ages in this normal distribution = 5.00

z: the z score = (x - ?)/?

?(z) = ?((x-?)/?): the probability that the ages in this normal distribution are less than x

Here is how it goes:

1.

z1 = (55.00 - 51.00)/5.00 = 0.8

z2 = (60.00 - 51.00)/5.00 = 1.8

P(x ? 60.00) = 0.96407

P(x ? 55.00) = 0.78814

P(55.00 ? x ? 60.00) = ?(1.80) - ?(0.80) = 0.964 - 0.788 = 0.176

2.

z1 = (48.00 - 51.00)/5.00 = -0.6

z2 = (52 - 51)/5 = 0.2

?(-0.60) = 0.274

?(0.20) = 0.579

P(48 ? x ? 52) = .........

3.

4.

You can do these now that I have shown it!

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.