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Use the Empirical Rule to answer each question.
During 1 week an overnight delivery company found that the weights of its parcels were normally distributed, with a mean of 28 ounces and a standard deviation of 7 ounces.
(a) What percent of the parcels weighed between 14 ounces and 35 ounces? (Round your answer to one decimal place.)
(b) What percent of the parcels weighed more than 49 ounces? (Round your answer to two decimal places.)

Sagot :

reinhard10158

Step-by-step explanation:

(a)

the probability of parcels weighing 35oz or less is

0.841

the probability of parcels weighing 14oz or less is

0.023

the probability of parcels weighing between 14 and 35oz is the probabilty of 35oz minus the probability of 14oz

0.841 - 0.023 = 0.818

the percentage is the probability × 100 = 81.8%

(b)

the probability of parcels weighing more than 49oz is 1 minus the probability to weigh less than 49oz.

the probability to weigh less than 49oz is

0.99865

so, the probability to weigh more than 49oz is

1 - 0.99865 = 0.00135

the percentage is again probability×100 = 0.14%

(a)The percent of the parcels weighed between 14 ounces and 35 ounces will be 0.818. (b) The percentage of the parcels weighing more than 49 ounces will be 0.14 %.

What is an empirical Rule?

It is the three-sigma rule, that asserts that with a normal distribution, virtually all observed data will lie within three standard deviations of the mean value.

(a)The percent of the parcels weighed between 14 ounces and 35 ounces will be 0.818.

The given data in the problem is;

The probability of parcels weighing 35 ounces or less, P₁=0.841

The probability of parcels weighing 14 ounces or less, P₂=0.023

The probability of parcels weighing between 14 and 35 ounces is found as;

P=P₁-P₂

P=0.841-01.023

P=0.818

The probability in the percentage will be;

[tex]\rm P = 0.818 \times 100 = 81.8 \%[/tex]

(b) The percent of the parcels weighing more than 49 ounces will be 0.14 %.

The probability of parcels weighing more than 49 ounces will be;

[tex]\rm P_3 = 1-0.99865 \\\\ \rm P_3 =0.00135 \\\\[/tex]

The probability in the percentage will be;

[tex]\rm P_3 = 0.00135 \times 100 \\\\ \rm P_3 =0.14 \%[/tex]

Hence the percentage of the parcels weighed between 14 ounces and 35 ounces will be 0.818.

To learn more about the empirical rule refer to the link;

https://brainly.com/question/11266206

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