Question

A stockbroker has kept a daily record of the value of a particular stock over the years and finds that prices of the stock form a normal distribution with a mean of $17.89 and a standard deviation of $5.27. The stock price beyond which 1% (01) of the distribution falls is- - --(7 points). 4. What is the value of r for the following relationship between X and Y (18 points)? 5. 87 17 19 86 21 59 36 23 25 47 A researcher collects data on the relationship between the amount of daily exercise an individual gets and the percent of body fat of the individual. The following scores are recorded. Calculate the least squares regression line (complete with a slope and intercept) predicting the % of fat () from amount of exercise (X) (18 points). 6. Individual 4 2 Exercise (min) 10182633 44 % Fat 30 2618 17 14

Answer 1

Answer)

4)

As the population is normally distributed, we can use standard normal z table to estimate the probability.

First we need to estimate the z score above which one percent lies or below which 99% lies

From z table, P(Z<2.33) = 0.99

Therefore, Z = 2.33

And we know that,

Z = (X-mean)/s.d

Given mean = 17.89

S.d = 5.27

2.33 = (X-17.89)/5.27

X = 30.1691

Therefore, the required value above which 1% lies

Is = 30.1691

5)

First we need to estimate the mean and standard deviation for x and y

For x:

Mean = 21

Sx = 3.1623

For y:

Mean = 63

Sy = 22.9456

r = -0.89580641647762

so substituting the values

Sx anVara

Numerator would be = (17-21)*(87-63) + (19-21)*(86-63)....

And denominator would be = 3.1623*22.9456*(5-1)

After solving

r = -0.8958