## Frequency Distribution

**Solution**

Interval | Frequency | Relative frequency | Percentages (%) | Cumulative frequency |

20 - 22 | 5 | 0.33 | 33.33 | 5 |

23 - 25 | 3 | 0.20 | 20.00 | 8 |

26 - 28 | 3 | 0.20 | 20.00 | 11 |

29 - 31 | 4 | 0.27 | 26.67 | 15 |

Total | 15 | 1.00 | 100.00 |

## Standard Deviation

X | ||

20 | -5.06667 | 25.67111 |

22 | -3.06667 | 9.404444 |

23 | -2.06667 | 4.271111 |

22 | -3.06667 | 9.404444 |

26 | 0.933333 | 0.871111 |

27 | 1.933333 | 3.737778 |

29 | 3.933333 | 15.47111 |

29 | 3.933333 | 15.47111 |

30 | 4.933333 | 24.33778 |

21 | -4.06667 | 16.53778 |

30 | 4.933333 | 24.33778 |

23 | -2.06667 | 4.271111 |

27 | 1.933333 | 3.737778 |

22 | -3.06667 | 9.404444 |

25 | -0.06667 | 0.004444 |

376 |
0 |
166.9333 |

Standard deviation = √((∑▒(x-x ̅ )^2 )/(n-1)) =√(166.9333/14) = 3.453087

Item 30 – 33 Probability distribution (Normal)

Normal Distribution

A public university investigated its admissions data for all students admitted to the university during the last 15 years (20,500 students). The university’s mean combined entry score was 1230, with a standard deviation of 115. The distribution was approximately normal. Use this information to answer the following questions. (5 points each)

What proportionof the students scored1250 or above?

μ=1230,σ=115

P(x≥1250)=P(Z≥(1250-1230)/115)

=1-P(Z≤(1250-1230)/115)

= 1-P(Z≤0.1739)

= 1-0.5690

0.431

What proportion of students scored between 1320 and 1500?

P(1320≤x≤1250)=P((1320-1230)/115≤Z≤(1500-1230)/115)

P((1320-1230)/115≤Z≤(1500-1230)/115) = P(Z≤Z (1500-1230)/115)-P(Z≤(1320-1230)/115)

= 0.2075

Your goal is to score in the top 7%, what is the minimum scoreyou can receive to achieve this goal?

P(Z≥(X-1230)/115)=0.07

(X-1230)/115=ϕ^(-1) (1-007)

(X-1230)/115=1.4758

X=1400

What scores do you need to be between 40% and 75%?

P(Z≤(X-1230)/115)=0.75

(X-1230)/115=ϕ^(-1) (0.75)

(X-1230)/115=0.6745

X=1308

Also,

P(Z≤(X-1230)/115)=0.40

(X-1230)/115=ϕ^(-1) (0.40)

(X-1230)/115=-0.2533

X=1201

Hence, between 1200 and 1310 is needed.

## Mean, Mode & Median

A marriage counselor is interested in examining the effectiveness of his newly developed intervention aimed at reducing the number of destructive arguments in marital relationships. But, first, he decided to assess the base rate of destructive arguments in untreated married couples. He asked 10 married couples to report how many arguments they had in the past week. Below is the data he got:

2, 1, 9, 0, 12, 0, 4, 2, 0, 9, 5, 0, 15, 2, 2, 5, 14

7. Compute the mode: ____________ (1 points)

Mode = 0 and 2

8. Compute the median: ____________ (1 points)

Median = 1.5

9. Compute the mean: _____________ (1 points)

Mean = 5

10. Given the relationships among the mean, median, mode, what is the shape of the distribution? (2 points)

**Short Answer Questions
**

11. Explain what is measured by the sign (+ or -) of a z-score and what is measured by its numerical value. (1.5 points)

The Z score tells us how many standard deviation we are away from the mean. The positive sign implies that the raw score is greater than the mean while the negative signs implies that the raw score is less than the mean.

12. Describe what happens to the mean, the standard deviation, and the shape of a distribution when all of the scores are transformed into z-scores. (1.5 points)

When all the scores are transformed into Z score, the mean becomes zero, the standard deviation becomes one and the shape of the distribution becomes symmetrical

**Extra Credit(up to 5 points)
**

What is the major difference between an experiment and non-experiment (1 sentence)? Provide an example of an experiment (1-2 sentences). Provide an example of a non-experiment (1-2 sentences).

Experiment simply means repeated trials for possible outcomes. The outcome is not certain. While non experiment requires no trial with certain outcomes.