is there a table or something you can add because I cant tell anything with the information you gave.
Answer:
Yes, both np and n(1-p) are ≥ 10
Mean = 0.12 ; Standard deviation = 0.02004
Yes. There is a less than 5% chance of this happening by random variation. 0.034839
Step-by-step explanation:
Given that :
p = 12% = 0.12 ;
Sample size, n = 263
np = 263 * 0.12 = 31.56
n(1 - p) = 263(1 - 0.12) = 263 * 0.88 = 231.44
According to the central limit theorem, distribution of sample proportion approximately follow normal distribution with mean of p = 0.12 and standard deviation sqrt(p*(1 - p)/n) = sqrt (0.12 *0.88)/n = sqrt(0.0004015) = 0.02004
Z = (x - mean) / standard deviation
x = 22 / 263 = 0.08365
Z = (0.08365 - 0.12) / 0.02004
Z = −1.813872
Z = - 1.814
P(Z < −1.814) = 0.034839 (Z probability calculator)
Yes, it is unusual
0.034 < 0.05 (Hence, There is a less than 5% chance of this happening by random variation.
Answer:
The storage capacity of the USB is far greater than Kilobytes, but less than a Terabyte.
Step-by-step explanation:
The USB storage device has a capacity of (8 - 256) GB. This implies that the USB has the capacity to store files size from 8 Gigabytes to 256 Gigabytes.
8 Gigabytes ≅ 8 000 000 000 bytes
256 Gigabytes ≅ 256 000 000 000 bytes
Kilobytes would not be an appropriate descriptive measurement of the storage capacity of the USB because 1 kilobyte ≅ 1 000 bytes. Thus;
8 Gigabytes ≅ 8 000 000 kilobytes
256 Gigabytes ≅ 256 000 000 Kilobytes
The storage capacity of the USB is far greater than Kilobytes.
Terabytes can not be used to describe the measurement f the USB because; 1 Terabyte ≅ 1 000 000 000 000 bytes
Thus, the storage capacity of the USB is less than 1 Terabyte.
18.
18 x 3 - 14 = 40
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