a) Let the mean concentration of particulate matter collected on filter paper is 45.6 ppm, with the standard deviation is 4.4 ppm. Find the probability that the mean concentration of particulate matter in a sample of 87 filter papers will be within 47 ppm of the guideline level. (4 marks) b) Rainfall data were collected for 12 months at Location A in mm, Jan 209, Feb 174, Mac 268, Apr 300, May 246, Jun 174, Jul 183, Aug 219, Sep 243, Oct 308, Nov 373, Dec 284. According to the Department of Irrigation and Drainage report, the average rainfall for the year of 2019 was 300mm. Find the T-score of this sample. c) Body temperature is known to have normal distribution among healthy adults. A researcher collected data from a random sample of 20 healthy adults. It was found that the sample temperatures have a mean of 88.4oF and a sample standard deviation of 0.3oF. i.

$a)$

$\mu=45.6\\\sigma=4.4\\n=87$

Let the random variable $X$ denote the concentration of particulate matter.

We find the probability,

$p(\bar x\lt 47)=p({\bar x-\mu\over {\sigma\over \sqrt{n}}}\lt {47-\mu\over{\sigma\sqrt{n}}})=p(Z\lt{47-45.6\over{4.4\over \sqrt{87}}})=p(Z\lt 2.97)=\phi(2.97)=0.9985$

The probability that the mean concentration of particulate matter in a sample of 87 filter papers will be within 47 ppm of the guideline level is 0.9985

$b)$

Let the random variable $X$ denote the amount of rainfall for the year

$\bar x={\sum x\over n}={2981\over 12}= 248.4167$

The sample variance is,

$s^2={\sum x^2-{(\sum x)^2\over n}\over n-1}={781721-740530.08\over11}=3744.628788$

The standard deviation is,

$\sqrt{s^2}=\sqrt{3744.628788}=61.1934$

The hypotheses tested are,

$H_0:\mu=300\\vs\\H_1:\mu\not=300$

The test statistic(t-score) is,

$t={\bar x-\mu\over{s\over \sqrt {n}}}={248.4167-300\over {61.1934\over \sqrt{12}}}=-2.92$

Therefore, the t score for this sample is -2.92

$c)$

Let the random variable $Y$ represent body temperature. We are given that,

$n=20\\\bar x=88.4\\s=0.3$

If we were to perform an hypothesis test for the mean, we would apply the t distribution where the number of degrees of freedom would be $n-1=20-1=19$

Therefore, the number of degrees of freedom for this test is 19.