25 POINTS!!!
2 answers:
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Complete Question
Suppose the weight of Chipotle burritos follows a normal distribution with mean of 450 grams, and variance of 100 grams2 . Define a random variable to be the weight of a randomly chosen burrito.
(a) What is the probability that a Chipotle burrito weighs less than 445 grams? (3 points)
(b) 20% of Chipotle burritos weigh more than what weight
Answer:
a
b
Step-by-step explanation:
From question we are told that
The population mean is
The variance is
The consider weight is
The standard deviation is mathematically represented as
substituting values
Given that weight of Chipotle burritos follows a normal distribution the the probability that a Chipotle burrito weighs less than x grams is mathematically represented as
Where is equal to z (the standardized values of the random number X )
So
substituting values
Now from the normal distribution table the value for is
=>
Let the probability of the Chipotle burritos weighting more that k be 20% so
=>
=>
From the normal distribution table the value of z for is
=>
=>
Finite approximation method of estimating the area under the curve of the
given function makes use of rectangular approximation of the area.
The correct responses are;
i) The estimated area using a lower sum with two rectangles of equal width is <u>1,715 square units</u>.
ii) The estimated area using a lower sum with four rectangles of equal width is <u>3,001.25 square units</u>.
iii) The estimated area using an upper sum with two rectangles of equal width is<u> 8,575 square units</u>.
iv) The estimated area using a upper sum with four rectangles of equal width is <u>6,431.25 square units</u>.
Reasons:
The given function is f(x) = 5·x²
The given domain is x₀ to x₁₄
i) Estimate using lower sum with two rectangles of equal width;
f(0) = 0
f(7) = 5 × 7² = 245
A = 0 × 7 + 245 × 7 = 1,715
The estimated area using a lower sum with two rectangles of equal width
is <u>1,715 square units</u>.
ii) Estimate using lower sum with four rectangles of equal width;
f(0) = 0
f(3.5) = 5 × 3.5² = 61.25
f(7) = 5 × 7² = 245
f(10.5) = 5 × 10.5² = 551.25
A = 0 × 3.5 + 61.25 × 3.5 + 245 × 3.5 + 551.25 × 3.5 = 3,001.25
The estimated area using a lower sum with four rectangles of equal width is <u>3,001.25 square units</u>.
iii) Estimate using an upper sum with two rectangles of equal width;
f(7) = 5 × 7² = 245
f(14) = 5 × 14² = 980
A = 245 × 7 + 980 × 7 = 8575
The estimated area using an upper sum with two rectangles of equal width
is <u>8,575 square units</u>.
iv) Estimate using an upper sum with four rectangles of equal width;
f(3.5) = 5 × 3.5² = 61.25
f(7) = 5 × 7² = 245
f(10.5) = 5 × 10.5² = 551.25
f(14) = 5 × 14² = 980
A = 61.25 × 3.5 + 245 × 3.5 + 551.25 × 3.5 + 980 × 3.5 = 6,431.25
The estimated area using a upper sum with four rectangles of equal width
is <u>6,431.25 square units</u>.
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