Question

Chucky grabbed 121212 items in the grocery store that each had a different price and had a mean cost of about \$7.41$7.41dollar sign, 7, point, 41. One of the items was an entire wheel of cheese that cost \$39.99$39.99dollar sign, 39, point, 99. [Show data]
\$1.29dollar sign, 1, point, 29
\$1.92dollar sign, 1, point, 92
\$3.19dollar sign, 3, point, 19
\$3.79dollar sign, 3, point, 79
\$3.99dollar sign, 3, point, 99
\$4.79dollar sign, 4, point, 79
\$5.19dollar sign, 5, point, 19
\$5.29dollar sign, 5, point, 29
\$5.49dollar sign, 5, point, 49
\$6.75dollar sign, 6, point, 75
\$7.19dollar sign, 7, point, 19
\$39.99dollar sign, 39, point, 99
Chucky then decided to put the wheel of cheese back and only buy the other 111111 items.
How will removing the wheel of cheese affect the mean and median?
Choose 1 answer:

Answer 1

Answer:

Mean reduces . New $mean(x)=4.443$

Median reduces. New $median(x)=4.79$

Step-by-step explanation:

Items(x)=1.29,1.92,3.19,3.79,3.99,4.79,5.19,5.29,5.49,6.75,7.19,39.99

Mean(x)

$\sum\frac{x_i}{n}\\=\sum x_i=1.29+1.92+3.19+3.79+3.99+4.79+5.19+5.29+5.49+6.75+7.19+39.99\\n=12\\mean(x)=\frac{88.92}{12}\\=7.41$

Removing last item(39.99)=>

$new \ sum=88.92-39.99\\=48.93\\New \ mean=\frac{48.93}{11}\\=4.44$

For median, median(x)=(n + 1) ÷ 2}th item (Arrange in numerical order)

$x_i$=1.29,1.92,3.19,3.79,3.99,4.79,5.19,5.29,5.49,6.75,7.19,39.99

=$\frac{4.79+5.19}{2}$=4.99

New Median:-

Since 11 is odd, the median item will be the 6th item.

1.29,1.92,3.19,3.79,3.99,4.79,5.19,5.29,5.49,6.75,7.19

New Median(x)=4.79