Question

Assume that females have pulse rates that are normally distributed with a mean of mu equals 72.0 beats per minute and a standard deviation of sigma equals 12.5 beats per minute. Complete parts​ (a) through​ (c) below. a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 79 beats per minute. The probability is 0.7123. ​(Round to four decimal places as​ needed.) b. If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 79 beats per minute. The probability is __.

Answer 1

Answer:

a) $P(X<79)=P(\frac{X-\mu}{\sigma}<\frac{79-\mu}{\sigma})=P(Z<\frac{79-72}{12.5})=P(Z<0.56)=0.7123$

b)$P(\bar X <79)=P(Z<\frac{79-72}{\frac{12.5}{\sqrt{25}}}=2.8)$

And using a calculator, excel ir the normal standard table we have that:

$P(Z<2.8)=0.9974$

c)  Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the female pulse rates of a population, and for this case we know the distribution for X is given by:

$X \sim N(72,12.5)$  

Where $\mu=72$ and $\sigma=12.5$

We are interested on this probability

$P(X<79)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X<79)=P(\frac{X-\mu}{\sigma}<\frac{79-\mu}{\sigma})=P(Z<\frac{79-72}{12.5})=P(Z<0.56)=0.7123$

Part b

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

The new z score is given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

$P(\bar X <79)=P(Z<\frac{79-72}{\frac{12.5}{\sqrt{25}}}=2.8)$

And using a calculator, excel ir the normal standard table we have that:

$P(Z<2.8)=0.9974$

Part c

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.