The table below is partially completed. Finish it and use it to find the mean and standard deviation. P(z) P(x) (x – µ) (2 – 4)° (x – µ)°P(æ) 0.14 -2.66 7.0756 0.990584 0.04 0.04 -1.66 2.7556 0.110224 0.24 0.48 -0.66 0.4356 0.104544 0.18 0.40 O Mean = 2.66 and Standard Deviation = 1.39 Mean = 0.52 and Standard Deviation = 1.39 Mean = 2.66 and Standard Deviation = 1.94 Mean = 0.52 and Standard Deviation = 1.94 Mean =-2.66 and Standard Deviation = 1.94 11234

MATLAB: An Introduction with Applications
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Author:Amos Gilat
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In a pizza takeout restaurant, the following probability distribution was obtained. The random variable z
represents the number of toppings for a large pizza.
The table below is partially completed. Finish it and use it to find the mean and standard deviation.
P(x) | xP(x)
(x – µ) | (r – µ)° (2 - 4)°P(@)
0.14
-2.66
7.0756
0.990584
0.04
0.04
-1.66
2.7556
0.110224
2
0.24
0.48
-0.66
0.4356
0.104544
3.
0.18
4
0.40
Mean = 2.66 and Standard Deviation = 1.39
Mean = 0.52 and Standard Deviation = 1.39
Mean = 2.66 and Standard Deviation = 1.94
Mean = 0.52 and Standard Deviation = 1.94
Mean = -2.66 and Standard Deviation = 1.94
Transcribed Image Text:In a pizza takeout restaurant, the following probability distribution was obtained. The random variable z represents the number of toppings for a large pizza. The table below is partially completed. Finish it and use it to find the mean and standard deviation. P(x) | xP(x) (x – µ) | (r – µ)° (2 - 4)°P(@) 0.14 -2.66 7.0756 0.990584 0.04 0.04 -1.66 2.7556 0.110224 2 0.24 0.48 -0.66 0.4356 0.104544 3. 0.18 4 0.40 Mean = 2.66 and Standard Deviation = 1.39 Mean = 0.52 and Standard Deviation = 1.39 Mean = 2.66 and Standard Deviation = 1.94 Mean = 0.52 and Standard Deviation = 1.94 Mean = -2.66 and Standard Deviation = 1.94
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