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mote1985 [20]
3 years ago
9

Average box of crackers is 24.5 ounces with standard deviation of. 8 ounce. What percent of the boxes weigh more than 22.9 ounce

s? What percent weigh less than 23.7 ounces?
Mathematics
1 answer:
34kurt3 years ago
7 0

Answer:

97.7% of of the boxes weigh more than 22.9 ounces.

15.9% of of the boxes weigh less than 23.7 ounces.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  24.5 ounces

Standard Deviation, σ = 0.8 ounce

We are given that the distribution of boxes weight is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(boxes weigh more than 22.9 ounces)

P(x > 22.9)

P( x > 22.9) = P( z > \displaystyle\frac{22.9 - 24.5}{0.8}) = P(z > -2)

= 1 - P(z \leq -2)

Calculation the value from standard normal z table, we have,  

P(x > 22.9) = 1 - 0.023 =0.977= 97.7\%

97.7% of of the boxes weigh more than 22.9 ounces.

b) P(boxes weigh less than 23.7 ounces)

P(x < 23.7)

P( x < 23.7) = P( z < \displaystyle\frac{23.7 - 24.5}{0.8}) = P(z < -1)

Calculation the value from standard normal z table, we have,  

P(x < 23.7) =0.159= 15.9\%

15.9% of of the boxes weigh less than 23.7 ounces.

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Estimate 1 13 cos(x2) dx 0 using the Trapezoidal Rule and the Midpoint Rule, each with n = 4. (Round your answers to six decimal
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(b) 3.215557

Step-by-step explanation:

(a)

\int\limits^{13}_1 {cos(x^2)} \, dx =M_n=$\sum_{n=1}^{\infty} f(m_i)\Delta x $

n=4, so :

Each subinterval has length :

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(b)

\int\limits^{13}_1 {cos(x^2)} \, dx =T_n=\frac{\Delta x}{2} (f(x_o)+2f(x_1)+2f(x_2)+...+2f(x_n_-_1)+f(x_n))

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Each subinterval has length :

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Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

The endpoints of the subintervals consist of:

5,9

Hence:

T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557

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