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notka56 [123]
3 years ago
9

Suzanne owns a small business that employs 555 other people. Suzanne makes \$100{,}000$100,000dollar sign, 100, comma, 000 per y

ear, and the other 555 employees make between \$40{,}000$40,000dollar sign, 40, comma, 000 and \$50{,}000$50,000dollar sign, 50, comma, 000 per year.
Suzanne decides to increase her salary by \$30{,}000$30,000dollar sign, 30, comma, 000 per year and leave the rest of the salaries the same.
How will increasing her salary affect the mean and median? How will increasing her salary affect the mean and median?
Mathematics
2 answers:
zalisa [80]3 years ago
6 0

Answer:

16

Step-by-step explanation:

Tpy6a [65]3 years ago
6 0

Answer:

Mean increased

Median remains same

Step-by-step explanation:

Suzanne make $ 100 k per year

5 other employee  $ 40 k  to $ 50 K per year per employee

data in ascending order

$40000  , $40000+a  , $40000+b  , $40000+c   , $50000  , $ 100000

where 10000 ≥c ≥ b ≥ a ≥ 0

Mean = ($40000 + $40000+a + $40000+b  + $40000+c  + $50000 + $ 100000) /6

Mean = $(310000 + a + b + c)/ 6

Median = ($40000+b  + $40000+c )/2 =  $40000 + (b+c)/2

Suzanne make now $ 100000 + $  30000 = $ 130000 per year

$40000  , $40000+a  , $40000+b  , $40000+c   , $50000 , $ 130000

Mean = ($40000 + $40000+a + $40000+b  + $40000+c  + $50000 + $ 130000) /6

Mean = $(340000 + a + b + c)/ 6

Median = ($40+b k + $40+c k)/2 =  $40 + (b+c)/2 k

Mean increased by  $(340000 + a + b + c)/ 6 - $(310000 + a + b + c)/ 6

= $ 30000/6  = $ 5000

Median Remains the same

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stiv31 [10]

Answer: 22.0.6%

Step-by-step explanation:

Given : According to a human modeling​ project, the distribution of foot lengths of women is approximately Normal with \mu=23.3\ cm and \sigma=1.3\ cm.

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P(x\leq22.4)=P(z\leq\dfrac{22.4-23.4}{1.3})\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\\approx P(z\leq-0.77)\\\\=1-P(z\leq0.77)\\\\=1-0.77935=0.2206499\approx0.2206=22.06\%

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