Question

A Sample Of 100 Clients Of An Exercise Facility Was Selected. Let X- The Number Of Days Per Week That A Randomly Selected Client Uses The Exercise Facility X | Frequency | 1 15 2 30 3 28 4 1O Find The Number That Is 1.5 Standard Deviations BELOw The Mean. (Round Your Answer To Three Decimal Places.)

Answer 1

Answer: 1.013

Step-by-step explanation:

Given the following data :

X_______f___fx___fx²

1_______15__15___15

2______30__60__120

3______28__84__252

4______10__40___160

______ 83__199__547

Number that is 1.5 standard deviations below the mean :

Mean(m) :

Σfx /Σf

(15 + 30 + 28 + 10) / 83

= 199 / 83

= 2.3976

Standard deviation(s) :

Sqrt[(Σfx² - (Σfx)²/Σf) /Σf-1]

sqrt[(547 - (199)²/83) / 83 - 1]

sqrt[(547 - (39601/83) / 82]

sqrt[(547 - 477.12) / 82]

sqrt(0.8522)

Standard deviation = 0.9231

1.5 standard deviations below the mean:

Mean - 1.5(standard deviation)

2.3976 - 1.5(0.9231)

2.3976 -1.38465

= 1.01295

= 1.013