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galina1969 [7]
3 years ago
7

Một lô hàng chứa rất nhiều sản phẩm, trong đó tỷ lệ sản phẩm loại tốt là 70%. chọn ngẫu nhiên từ lô hàng ra 5 sản phẩm. gọi X là

số sản phẩm tốt có trong 5 sản phẩm chọn ra.
a. hãy tìm bảng phân phối xác xuất của X
b. xác định kỳ vọng và phương sai của X
c. tìm số sản phẩm tốt được lấy ra có khả năng xảy ra nhiều nhất
Mathematics
1 answer:
den301095 [7]3 years ago
8 0

trả lời : lâp bảng

X=0 thì xác suất lấy ra được sản phẩm tốt là 5*30%=1,5%

X =1 thì xác suất lấy ra được là 1* 70%*4*30%=0,84%

X=2 thì xác suất lấy ra được sản phẩm tốt là 2*70%*3*30%=1,26%

X=3 thì xác suất lấy ra được sản phẩm tốt là 3*70%*2*30%=1,26%

X=4 thì xác suất lấy ra được sản phẩm tốt là 4*70%*1*30%=0,84%

X=5 thì xác suất lấy ra được sản phẩm tốt là 5*70%=3,5%

X 0 1 2 3 4 5

P 1,5 0,84 1,26 1,26 0,84 3,5

b,

E(x)= 1,5*0+ 0,84*1+1,26*2+1,26*3+0,84*4+3,5*5=28

E(x^2)=1,5*0^2+0,84*1^2+1,26*2^2+1,26*3^2+0,84*4^2+3,5*5^2=118,16

===> V(x)= (E(x))^2-E(x^2)=28^2-118,16=665,84

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The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

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Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

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From the data we have;

The mean waiting time before the downsize, \overline x_1 = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, \overline x_2 = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20  - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; t_{\alpha /2} = ±2.026192

The t-confidence interval is given as follows;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore;

\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

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