STATISTICAL ANALYSIS

Statistical Analysis

Statistical Analysis

Solution to 3.13

Statistics

Chocolate Chip Cookies

N

Valid

13

Missing

0

Mean

33.692

Median

36.000

Mode

25.0 a

Std. Deviation

13.5732

Variance

184.231

Skewness

-.361

Std. Error of Skewness

.616

Range

47.0

Sum

438.0

a. Multiple modes exist. The smallest value is shown

Coefficient of Variation

=

SD

=

13.5732

=

0.403

x

33.692

The given data pertaining to Chocolate Chip Cookies is skewed as the mean and median are not equal.

With reference to the cost of chocolate chip cookies, it can be noted that the data is negatively skewed as the value of skewness is -0.36; besides it, it is also observed that there is too much variation in the data as the standard deviation also the variance of the Chocolate Chip Cookies is high that are 13.5 and 184 respectively.

Solution to 3.63

Statistics

Number of days between the resolution of the complaint and the receipt of a complaint

N

Valid

50

Missing

0

Mean

43.0400

Median

28.5000

Mode

5.00a

Std. Deviation

41.93

Variance

1758

Skewness

1.488

Std. Error of Skewness

.337

Range

164.00

Sum

2152.00

a. Multiple modes exist. The smallest value is shown

Coefficient of Variation

=

SD

=

41.93

=

0.974

x

43.04

The given data relating to receipt and resolution of a complaint is skewed as the mean and median are not equal.

In relation to then number of days between the resolution of the complaint and the receipt of a complaint, it can be said that the data is positively skewed as the value of skewness is 1.48. In addition to this, it is also observed that there is deviation in the data as the standard deviation also the variance of the receipt and resolution of a complaint varies too much.

Solution to 4.13

(a) Order at the drive - through = 138/ 200 = 0.69

(b) Male and prefers to order at the drive - through = 60/ 100 = 0.6

(c) Male or prefers to order at the drive - through = 138/ 200 = 0.69

(d) From the above part b and c, it can be observed that in part b is showing the probability of male to order at the drive through; however, the part c is presenting the probability of both females and male to order at the drive through.

Solution to 4.23

(a) Probability of male prefers to order at the drive - through = 60/ 100 = 0.6

(b) Probability of female prefers to order at the drive - through = 78/ 100 = 0.78

(c) Dining preference independent of gender;

Male = 21/ 100 = 0.21

Female = 12/ 100 = 0.12

It can be said that the dining preference is dependent of gender as the male often prefer to dine as compared to females.

Solution to 5.27

(a)

P (X = 3|n = 3, p = 0.901) =

= 1 (0.901) (0.901) (0.901) (1) = 0.731432701

(b)

P (X = 3|n = 3, p = 0.901) =

= 1 (1) (0.099) (0.099) (0.099) = 0.000970299

(c)

P (X = 3|n = 3, p = 0.901) =

= 3 * 0.901 * 0.901 * 0.099 = 0.241104897

P (x>= 2) = P (x = 2) + P (x = 3)

P (x>= 2) = 0.241104897 + 0.731432701

P (x>= 2) = 0.972537598

(d)

µ = 3 * 0.901 = 2.703

s = v 3 * 0.901 * 0.099 = v 0.267 = 0.517

(e)

The mean number of orders filled correctly in a sample of three orders is 2.7, and the standard deviation is ...