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Use the empirical rule (68-95-99.7) to determine if the following maximum daily rainfall totals could be considered to be normally distributed. (16 points) 1.1, 1.2, 1.5, 1.55, 1.71, 1.75, 1.9, 1.94, 1.99, 2.0, 2.05, 2.08, 2.12, 2.18, 2.2, 2.25, 2.36, 2.5, 2.61, 2.81, 2.88, 2.95, 3.0, 3.03, 3.07, 3.16, 3.25, 3.62.

User AnthonyI
by
5.1k points

1 Answer

5 votes

Answer:

Explanation:

Given that:

The number of observations (n) = 28

The mean
\overline x is the addition of all the given numbers divided by the total number of observations = 64.76/28


\overline x = 2.31

The standard deviation s = 0.65

At 68%; the empirical rule falls between:

=
\overline x \pm s

= 2.31 ± 0,65

= ( 2.31 - 0.65, 2.31 + 0.65)

= (1.66 , 2.96)

From the parameters given:

4 values are < 1.66 and 6 values are > 2.96

Thus, the percentage covered = (28 - (6+4))/28 × 100

= (28 - 10) / 28 × 100

= (18/28) × 100

= 64.29%

Hence, this satisfies the normal distribution

At 95; the empirical rule falls between:

=
\overline x \pm 2 s

= 2.31 ± 2(0.65)

= 2.31 ± 1.30

= ( 2.31 - 1.30, 2.31 + 1.30)

= (1.01 , 3.61)

From the parameters given:

0 values are < 1.01 and 1 values are > 3.61

Thus, the percentage covered = (28 - (1+0))/28 × 100

= (28 - 1) / 28 × 100

= (27/28) × 100

= 96.43%

Hence, this satisfies the normal distribution

At 99.7; the empirical rule falls between:

=
\overline x \pm 3 s

= 2.31 ± 3(0.65)

= 2.31 ± 1.95

= ( 2.31 - 1.95, 2.31 + 1.95)

= (0.36 , 4.26)

From the parameters given:

0 values are < 0.36 and 0 values are > 4.26

Thus, the percentage covered = (28 - (0+0))/28 × 100

= (28 - 0) / 28 × 100

= (28/28) × 100

=100%

Hence, this satisfies the normal distribution and the maximum daily rainfall totals could be considered to be normally distributed.

User Seyed Mohammad
by
5.8k points
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