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goldfiish [28.3K]
2 years ago
6

Thời gian vận chuyển hàng hóa trên tuyến đường AB là biến ngẫu nhiên X có phân phối

Mathematics
2 answers:
steposvetlana [31]2 years ago
8 0

Answer:

Step-by-step explanation:

P(X>5.5)= 0.5-phi((5.5-5)/0.25)

=0.5- phi(2)

= 0.5-0.4773= 0.0227

Goshia [24]2 years ago
4 0

Answer:

Thời gian vận chuyển hàng hóa trên tuyến đường AB là biến ngẫu nhiên X có phân phối

chuẩn với trung bình   5 giờ, độ lệch chuẩn   0, 25 giờ.

a. Chuyến có thời gian lớn hơn 5,2 ( ) /100   a b  giờ được xem là trễ, tính tỉ lệ chuyến trễ?

Step-by-step explanation:

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Answer:

a.  X~N(2,885, 651)

b.  0.086291

c.  0.00058

d.  3213.10 calories

Step-by-step explanation:

a. -A normal distribution is expressed in the form X~N(mean, standard deviation).

-Let X a random variable denoting  the number of calories consumed.

-X is a is a normally distributed random variable with mean 2885 and standard deviation 651.

-This distribution is expressed as X~N(2,885, 651)

b. The probability that less than 2000 calories are consumed is calculated using the formula:

P(X

#substitute the given values in the formula to solve for P:

P(X

Hence, the probability of consuming less than 2000 calories is 0.08691

c. The proportion of customers consuming more than 5000 calories is calculated as:

P(X>x)=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(Z>\frac{5000-2885}{651})\\\\=P(z>3.2488)\\\\=1-0.99942\\\\=0.00058

Hence, the proportion of customers consuming over  5000 calories is 0.00058

d. The least amount of calories to get the award is calculated as:

1% is equivalent to a z value of 0.50399.

-We equate this to the formula to solve for the mean consumption:

0.01=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(z>\frac{\bar X-2885}{651})\\\\\1\%=0.50399 \\\\\frac{\bar X-2885}{651}=0.50399 \\\\\bar X=0.50399\times 651+2885\\\\=3213.09

Hence, the least amount of calories consumed to qualify for the award is 3213.10 calories.

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