To solve this problem, we make use of the z statistic. The formula for the z score is:z score = (x – u) / swhere x is the sample value = 0.90, u is the sample mean = 0.917, and s is the standard deviation = 0.005
Therefore:z score = (0.90 – 0.917) / 0.005z score = -3.4
From the standard probability tables, the p-value for a right tailed test of z = -3.4 is:P = 0.9997
Therefore there is a 99.97% chance that it will be above 0.90 mm
Answer:
For a data from population which is not normally distributed, the sample means would be approximately a normal distribution if the sample size (n) is greater than 30
Step-by-step explanation:
For a data from population which is not normally distributed, the sample means would be approximately a normal distribution if the sample size (n) is greater than 30 i.e n ≥ 30 this is because the shape of a sample distribution depends on the sample size. But for normal distribution population, the sample means would be approximately a normal distribution even if the sample size is less than 30;
Answer:
373,952 dollars
Step-by-step explanation:
375/1,0028
Multiply the first equation by two to get 2x + 2y = 4.
Subtract the second equation from the first equation to get -3y = -12
y = 4.
Plug y back in to get x = -2.