Question

Could someone pls help with this maths homework?
The age of members of a curling club is normally
distributed, with a mean of 63 years and a standard deviation of 4 years. What
percentage of male and female curlers are there in each of the
following age groups?
a) 55 to 63 b) 67 to 75 c) over 75

Thank you for your time!!

Answer 1

Using the normal distribution, it is found that the percentages are as follows:

a) 47.5%.

b) 15.87%.

c) 0.13%.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem, the mean and the standard deviation are given by, respectively, $\mu = 63, \sigma = 4$.

Item a:

The proportion is the p-value of Z when X = 63 subtracted by the p-value of Z when X = 55, hence:

X = 63:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{63 - 63}{4}$

$Z = 0$

$Z = 0$ has a p-value of 0.5.

X = 55:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55 - 63}{4}$

$Z = -2$

$Z = -2$ has a p-value of 0.025.

0.5 - 0.025 = 0.475 = 47.5%.

Item b:

The proportion is the p-value of Z when X = 75 subtracted by the p-value of Z when X = 67, hence:

X = 75:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{75 - 63}{4}$

$Z = 3$

$Z = 3$ has a p-value of 0.9987.

X = 67:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 63}{4}$

$Z = 1$

$Z = 1$ has a p-value of 0.84.

0.9987 - 0.84 = 0.1587 = 15.87%.

Item c:

The proportion is one subtracted by the p-value of Z when X = 75, hence 1 - 0.9987 = 0.0013 = 0.13%.

More can be learned about the normal distribution at https://brainly.com/question/24663213