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klemol [59]
3 years ago
15

A produce supplier ships boxes of produce to individual customers. The distribution of weights of shipped boxes is approximately

normal with mean 36 pounds and standard deviation 4 pounds. Which expression represents the weight, in pounds, at the 75th percentile of the distribution?
(A) -1.96(4)+ 36
(B) -0.25(4) + 36
(C) 0.25(4)+ 36
(D) 0.67(4)+ 36
(E) 0.75(4)+ 36
Mathematics
2 answers:
Romashka [77]3 years ago
6 0

Answer:

Option D.

Step-by-step explanation:

Given information: The distribution of weights of shipped boxes is approximately normal.

Population mean = 36

Standard deviation = 4

We need to find the 75th percentile of the distribution

75% = 0.75

From z-table it is clear that the value of z at 0.75 is 0.674.

75th percentile = Mean + z value at 75%(Standard deviation)

75th percentile = 36 + 0.674(4)

In can be rewritten as

75th percentile = 0.67(4) + 36

Therefore, the correct option is D.

schepotkina [342]3 years ago
5 0

Answer:

0.67(4)+ 36

Option D

Step-by-step explanation:

Given that a producer ships boxes of produce to individual customers.(say x)

X is a normal random variable with mean = 36 and std dev  =4 lbs

By definition of std normal variate we know that

\frac{x-36}{4} is N(0,1)

75th percentile of std normal distribution is

0.675

Hence corresponding x would be

36+0.675(4)

Option D matches with this value

Hence answer is option D

0.67(4)+ 36

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Slav-nsk [51]
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Please someone help me to prove this. ​
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<h3><u>Answer</u> :</h3>

We know that,

\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}

\dag\bf\:sin2A=2sinA\:cosA

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

<u>Now, Let's solve</u> !

\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}

\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}

\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}

\leadsto\:\sf\dfrac{2sin\frac{2A+2B}{2}\:sin\frac{2A-2B}{2}}{2sin\frac{2A-2B}{2}\:cos\frac{2A+2B}{2}}

\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}

\leadsto\:\bf{tan(A+B)}

5 0
3 years ago
Read 2 more answers
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