3 - JustAnswer



3.     

a. What score was earned by more students than any other score? Why? 84. It's the mode.

b. How many students scored between 68 and 94 on the exam? 25 students. Because 50% of the students lie between the 1st and 3rd quartiles and there are 50 students.

c. What was the highest score earned on the exam? 98 (see below for calculations

(a+b)/2 = 72 (that's the midrange)

b-a = 52 (that's the range)

So we can solve these equations: b= 52+a

(2a+52)/2 = 72 --> a+26 = 72 --> a=46 --> b=98.

d. What was the lowest score earned on the exam? 46 (see above for calculations)

e. How many students scored within three standard deviations of the mean ? By Chebyshev's Theorem, at least 1-1/9 = 0.889% or 44.44 students scored within that interval.

Show your calculations leading to your standard deviation and variance on #2.

2.     A math test was given with the following results:

80, 69, 92, 75, 88, 37, 98, 92, 90, 81, 32, 50, 59, 66, 67, 66

Find the range, standard deviation, and variance for the scores.

To find the range we subtract the minimum number from the maximum number:

32, 37, 50, 59, 66, 66, 67, 69, 75, 80, 81, 88, 90, 92, 92, 98

Range = 98-32 = 66

Finding the variance, it’s helpful to use a table:

|  |x |x-(x-bar) |(x-(x-bar)^2) |  |

|  |32 |-39.375 |1550.391 |  |

|  |37 |-34.375 |1181.641 |  |

|  |50 |-21.375 |456.891 |  |

|  |59 |-12.375 |153.141 |  |

|  |66 |-5.375 |28.891 |  |

|  |66 |-5.375 |28.891 |  |

|  |67 |-4.375 |19.141 |  |

|  |69 |-2.375 |5.641 |  |

|  |75 |3.625 |13.141 |  |

|  |80 |8.625 |74.391 |  |

|  |81 |9.625 |92.641 |  |

|  |88 |16.625 |276.391 |  |

|  |90 |18.625 |346.891 |  |

|  |92 |20.625 |425.391 |  |

|  |92 |20.625 |425.391 |  |

|  |98 |26.625 |708.891 |  |

|x-bar |71.375 |  |5787.75 |0.533)=0.297

So, the proportion of male college students whose height is greater than 70 inches of 0.297

c.     Describe the distribution of , the mean of samples of size 16.

The sample average is normally distributed with mean of 68 inches and standard deviation of 3.75/sqrt(16) = 0.9375

d.     Find the mean and standard error of the distribution.

The sample average is normally distributed with mean of 68 inches and standard deviation of 3.75/sqrt(16) = 0.9375

e.     Find P (x-bar > 70) = 0.143 (see work below):

z=(70-68)/0.9375 = 2.133

P(z>2.133) = 0.016

f.     Find P (x-bar < 67) = 0.297 (see work below):

z=(67-68)/0.9375 = -1.066

P(z82.5) = 0.001349898

8. Find the value of z such that 40% of the distribution lies between it and the mean.

Let’s find the two values a and b such that

P(z ................
................

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