Question

Jessica grows two types of tomato plants in her garden. For each type, she records the heights, in inches, of five plants.

Answer 1

Answer:

(A)

Step-by-step explanation:

For type 1, the given data set is:

{42, 56, 44, 60, 48}

Mean of the given data is:

$Mean=\frac{42+56+44+60+48}{5}=50$

Data set                                                       $(x-{\overline{x}})^2$

42                                                                     64

56                                                                     36

44                                                                      36

60                                                                     100

48                                                                       4

Now, $Variance=\frac{64+36+36+100+4}{5}=48$

Thus, $standard deviation=\sqrt{Variance}=\sqrt{48}=6.92$

The difference in the standard deviation and mean of type 1 data set is =$50-6.92=43.08$.

For type 2:

{62, 59, 68, 55, 66}

Mean of the given data is:

$Mean=\frac{62+59+68+55+66}{5}=62$

Data set                                                       $(x-{\overline{x}})^2$

62                                                                     0

59                                                                     9

68                                                                    36

55                                                                    49

66                                                                     16

Now, $Variance=\frac{0+9+36+49+16}{5}=22$

Thus, $standard deviation=\sqrt{Variance}=\sqrt{22}=4.69$

The difference in the standard deviation and mean of type 1 data set is =$62-4.69=57.31$.

Since, difference in the standard deviation and mean of both type 1 and type 2 are different, therefore Type 1 plants and Type 2 plants do not have similar height distributions.

Answer 2

The answer is D 
Type 1 Type 2 plants have somewhat similar height distributions