Question

In certain drug trial, 10 subjects who received a placebo reported headaches, while 30 subjects who received a placebo reported no headaches. Of the subjects taking a new drug, 20 reported headaches, while 30 did not. a) Display this information in a contingency table, including all totals. b) What percentage of participants in the trial reported headaches? c) What percentage of new-drug takers reported headaches? d) What percentage of placebo takers reported headaches?

a) Draw a diagram to show a standard Normal distribution and shade in the regions between (µ–3(sigma)) and (µ+(sigma)). Find the approximate percentage of the population that would have values in this region. b) A researcher has found that among many mice in given maze the average time to complete the circuit is 10 minutes. The times were normally distributed with a standard deviation of 3 minutes. Find: i) the approximate proportion of mice that completed the maze in less than 4 minutes, ii) the probability that a random mouse takes more than 15 minutes to complete the maze, iii) the time to complete the maze below which are the fastest 5% of mice, and iv) the first quartile of the times to complete the maze

Use the following stem-and-leaf display of the ages of 10 persons.

1 | 2 4 4

2| 1 1 2 6 9

3| 2

4|

5 | 8

a) Find the mean and mode of the ages. b) By hand, find the five-number summary of the ages. c) Find the range and interquartile range of the ages. d) Provide a dot plot of the ages.

Answer 1

1)

a)

The contingency table is shown below,

	Placebo	New drug	Total
Headaches	10	20	30
No headaches	30	30	60
Total	40	50	90

b)

c)

d)

Next question

a)

For standard normal curve,

mean = 0,

Standard deviation = 1

The area between the value -3 and +3 is the regions between (µ–3(sigma)) and (µ+(sigma))

The percentage of the population is obtained as follow,

The probability for the Z value is obtained from normal distribution table,

b)

Mean = 10 minutes

Standard deviation = 3 minutes

i)

The probability is obtained by calculating the z score,

The probability for the Z value is obtained from normal distribution table,

ii)

The probability for the Z value is obtained from normal distribution table,

iii)

The z score for the probability = 0.05 is obtained from standard normal distribution table,

iv)

The first quartile is the 25th percentile,

The z score for the probability = 0.25 is obtained from standard normal distribution table,

Next question

a)

From the stem-and-leaf plot, the data values are,

Age
12
14
14
21
21
22
26
29
32
58

a)

Mean is the average value of the ages

Mode is the most frequent value in the data points,

b)

First sort the data values smallest to largest,

Age
12		Minimum = 12
14
14		First quartile = 14
21
21	Average of 21 and 22 = 21.5	Median = 21.5
22
26
29		Third quartile = 29
32
58		Maximum = 58

c)

IQR = Q3 - Q1 = 29 - 14 = 15