Solved on Mar 06, 2024

Rainfall in a region is normally distributed with $\mu=42.3$ inches, $\sigma=5.6$ inches. Find percentages of years with rainfall: a) $<44$ inches b) $>39$ inches c) between 38 and 43 inches.

STEP 1

Assumptions
1. The annual rainfall is normally distributed.
2. The mean (average) annual rainfall is 42.3 inches.
3. The standard deviation of the annual rainfall is 5.6 inches.

STEP 2

To find the percentage of years with less than 44 inches of rainfall, we will calculate the z-score for 44 inches. The z-score is the number of standard deviations a data point is from the mean.
$z = \frac{X - \mu}{\sigma}$
where $X$ is the value of interest (44 inches), $\mu$ is the mean, and $\sigma$ is the standard deviation.

STEP 3

Calculate the z-score for 44 inches.
$z = \frac{44 - 42.3}{5.6}$

STEP 4

Perform the calculation to find the z-score.
$z = \frac{44 - 42.3}{5.6} = \frac{1.7}{5.6} \approx 0.3036$

STEP 5

Using the z-score, we can now look up the corresponding percentile in the standard normal distribution table or use a calculator with a normal distribution function to find the percentage of years with less than 44 inches of rainfall.

STEP 6

Find the percentile corresponding to a z-score of 0.3036.
The percentile for z = 0.3036 is approximately 0.6187 or 61.87%.

STEP 7

Thus, about 61.87% of the years will have an annual rainfall of less than 44 inches.

STEP 8

To find the percentage of years with more than 39 inches of rainfall, we will calculate the z-score for 39 inches using the same formula as before.

STEP 9

Calculate the z-score for 39 inches.
$z = \frac{39 - 42.3}{5.6}$

STEP 10

Perform the calculation to find the z-score.
$z = \frac{39 - 42.3}{5.6} = \frac{-3.3}{5.6} \approx -0.5893$

STEP 11

We will now look up the corresponding percentile for a z-score of -0.5893.

STEP 12

Find the percentile corresponding to a z-score of -0.5893.
The percentile for z = -0.5893 is approximately 0.2776 or 27.76%.

STEP 13

To find the percentage of years with more than 39 inches of rainfall, we subtract the percentile from 1 (or 100%) because we want the area to the right of the z-score.
$Percentage > 39\ inches = 1 - 0.2776$

STEP 14

Perform the calculation to find the percentage.
$Percentage > 39\ inches = 1 - 0.2776 = 0.7224 \text{ or } 72.24\%$

STEP 15

Thus, about 72.24% of the years will have an annual rainfall of more than 39 inches.

STEP 16

To find the percentage of years with annual rainfall between 38 inches and 43 inches, we will calculate the z-scores for both 38 inches and 43 inches.

STEP 17

Calculate the z-score for 38 inches.
$z = \frac{38 - 42.3}{5.6}$

STEP 18

Perform the calculation to find the z-score for 38 inches.
$z = \frac{38 - 42.3}{5.6} = \frac{-4.3}{5.6} \approx -0.7679$

STEP 19

Calculate the z-score for 43 inches.
$z = \frac{43 - 42.3}{5.6}$

STEP 20

Perform the calculation to find the z-score for 43 inches.
$z = \frac{43 - 42.3}{5.6} = \frac{0.7}{5.6} \approx 0.1250$

STEP 21

Now we will look up the corresponding percentiles for both z-scores.

STEP 22

Find the percentile corresponding to a z-score of -0.7679.
The percentile for z = -0.7679 is approximately 0.2212 or 22.12%.

STEP 23

Find the percentile corresponding to a z-score of 0.1250.
The percentile for z = 0.1250 is approximately 0.5500 or 55.00%.

STEP 24

To find the percentage of years with annual rainfall between 38 inches and 43 inches, we subtract the smaller percentile from the larger one.
$Percentage\ between\ 38\ and\ 43\ inches = 0.5500 - 0.2212$

STEP 25

Perform the calculation to find the percentage.
$Percentage\ between\ 38\ and\ 43\ inches = 0.5500 - 0.2212 = 0.3288 \text{ or } 32.88\%$

STEP 26

Thus, about 32.88% of the years will have an annual rainfall between 38 inches and 43 inches.
The solutions to the problems are: a) Approximately 61.87% of years will have an annual rainfall of less than 44 inches. b) Approximately 72.24% of years will have an annual rainfall of more than 39 inches. c) Approximately 32.88% of years will have an annual rainfall between 38 inches and 43 inches.

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Rainfall in a region is normally distributed with μ=42.3\mu=42.3μ=42.3 inches, σ=5.6\sigma=5.6σ=5.6 inches. Find percentages of years with rainfall: a) <44<44<44 inches b) >39>39>39 inches c) between 38 and 43 inches.

STEP 1

Assumptions1. The annual rainfall is normally distributed.2. The mean (average) annual rainfall is 42.3 inches.3. The standard deviation of the annual rainfall is 5.6 inches.

STEP 2

STEP 3

Calculate the z-score for 44 inches.z=44−42.35.6 z = \frac{44 - 42.3}{5.6} z=5.644−42.3​

STEP 4

Perform the calculation to find the z-score.z=44−42.35.6=1.75.6≈0.3036 z = \frac{44 - 42.3}{5.6} = \frac{1.7}{5.6} \approx 0.3036 z=5.644−42.3​=5.61.7​≈0.3036

STEP 5

Using the z-score, we can now look up the corresponding percentile in the standard normal distribution table or use a calculator with a normal distribution function to find the percentage of years with less than 44 inches of rainfall.

STEP 6

Find the percentile corresponding to a z-score of 0.3036.The percentile for z = 0.3036 is approximately 0.6187 or 61.87%.

STEP 7

Thus, about 61.87% of the years will have an annual rainfall of less than 44 inches.

STEP 8

To find the percentage of years with more than 39 inches of rainfall, we will calculate the z-score for 39 inches using the same formula as before.

STEP 9

Calculate the z-score for 39 inches.z=39−42.35.6 z = \frac{39 - 42.3}{5.6} z=5.639−42.3​

STEP 10

Perform the calculation to find the z-score.z=39−42.35.6=−3.35.6≈−0.5893 z = \frac{39 - 42.3}{5.6} = \frac{-3.3}{5.6} \approx -0.5893 z=5.639−42.3​=5.6−3.3​≈−0.5893

STEP 11

We will now look up the corresponding percentile for a z-score of -0.5893.

STEP 12

Find the percentile corresponding to a z-score of -0.5893.The percentile for z = -0.5893 is approximately 0.2776 or 27.76%.

STEP 13

To find the percentage of years with more than 39 inches of rainfall, we subtract the percentile from 1 (or 100%) because we want the area to the right of the z-score.Percentage>39 inches=1−0.2776 Percentage > 39\ inches = 1 - 0.2776 Percentage>39 inches=1−0.2776

STEP 14

Perform the calculation to find the percentage.Percentage>39 inches=1−0.2776=0.7224 or 72.24% Percentage > 39\ inches = 1 - 0.2776 = 0.7224 \text{ or } 72.24\% Percentage>39 inches=1−0.2776=0.7224 or 72.24%

STEP 15

Thus, about 72.24% of the years will have an annual rainfall of more than 39 inches.

STEP 16

To find the percentage of years with annual rainfall between 38 inches and 43 inches, we will calculate the z-scores for both 38 inches and 43 inches.

STEP 17

Calculate the z-score for 38 inches.z=38−42.35.6 z = \frac{38 - 42.3}{5.6} z=5.638−42.3​

STEP 18

Perform the calculation to find the z-score for 38 inches.z=38−42.35.6=−4.35.6≈−0.7679 z = \frac{38 - 42.3}{5.6} = \frac{-4.3}{5.6} \approx -0.7679 z=5.638−42.3​=5.6−4.3​≈−0.7679

STEP 19

Calculate the z-score for 43 inches.z=43−42.35.6 z = \frac{43 - 42.3}{5.6} z=5.643−42.3​

STEP 20

Perform the calculation to find the z-score for 43 inches.z=43−42.35.6=0.75.6≈0.1250 z = \frac{43 - 42.3}{5.6} = \frac{0.7}{5.6} \approx 0.1250 z=5.643−42.3​=5.60.7​≈0.1250

STEP 21

Now we will look up the corresponding percentiles for both z-scores.

STEP 22

Find the percentile corresponding to a z-score of -0.7679.The percentile for z = -0.7679 is approximately 0.2212 or 22.12%.

STEP 23

Find the percentile corresponding to a z-score of 0.1250.The percentile for z = 0.1250 is approximately 0.5500 or 55.00%.

STEP 24

STEP 25

Perform the calculation to find the percentage.Percentage between 38 and 43 inches=0.5500−0.2212=0.3288 or 32.88% Percentage\ between\ 38\ and\ 43\ inches = 0.5500 - 0.2212 = 0.3288 \text{ or } 32.88\% Percentage between 38 and 43 inches=0.5500−0.2212=0.3288 or 32.88%

STEP 26

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Rainfall in a region is normally distributed with $\mu=42.3$ inches, $\sigma=5.6$ inches. Find percentages of years with rainfall: a) $<44$ inches b) $>39$ inches c) between 38 and 43 inches.

Assumptions
1. The annual rainfall is normally distributed.
2. The mean (average) annual rainfall is 42.3 inches.
3. The standard deviation of the annual rainfall is 5.6 inches.

Calculate the z-score for 44 inches.
$z = \frac{44 - 42.3}{5.6}$

Perform the calculation to find the z-score.
$z = \frac{44 - 42.3}{5.6} = \frac{1.7}{5.6} \approx 0.3036$

Find the percentile corresponding to a z-score of 0.3036.
The percentile for z = 0.3036 is approximately 0.6187 or 61.87%.

Calculate the z-score for 39 inches.
$z = \frac{39 - 42.3}{5.6}$

Perform the calculation to find the z-score.
$z = \frac{39 - 42.3}{5.6} = \frac{-3.3}{5.6} \approx -0.5893$

Find the percentile corresponding to a z-score of -0.5893.
The percentile for z = -0.5893 is approximately 0.2776 or 27.76%.

To find the percentage of years with more than 39 inches of rainfall, we subtract the percentile from 1 (or 100%) because we want the area to the right of the z-score.
$Percentage > 39\ inches = 1 - 0.2776$

Perform the calculation to find the percentage.
$Percentage > 39\ inches = 1 - 0.2776 = 0.7224 \text{ or } 72.24\%$

Calculate the z-score for 38 inches.
$z = \frac{38 - 42.3}{5.6}$

Perform the calculation to find the z-score for 38 inches.
$z = \frac{38 - 42.3}{5.6} = \frac{-4.3}{5.6} \approx -0.7679$

Calculate the z-score for 43 inches.
$z = \frac{43 - 42.3}{5.6}$

Perform the calculation to find the z-score for 43 inches.
$z = \frac{43 - 42.3}{5.6} = \frac{0.7}{5.6} \approx 0.1250$

Find the percentile corresponding to a z-score of -0.7679.
The percentile for z = -0.7679 is approximately 0.2212 or 22.12%.

Find the percentile corresponding to a z-score of 0.1250.
The percentile for z = 0.1250 is approximately 0.5500 or 55.00%.

Perform the calculation to find the percentage.
$Percentage\ between\ 38\ and\ 43\ inches = 0.5500 - 0.2212 = 0.3288 \text{ or } 32.88\%$