Asurvey of 9,000 customers at ron’s grocery store showed that 35% (3,150 customers) would buy grainy-os cereal. in the first month, 4,200 out of the 9,000 customers bought the cereal. what was the percent error in the survey estimate of the number of people buying grainy-os?

The percent error in the survey is 25%

Step-by-step explanation:

- Percent error is the difference between an approximated value and

 exact value, divided by the exact value, multiplied by 100%

- Percent of error is a positive value

- error% = [I approximated value - exact value I ÷ exact value] × 100%

- A survey of 9,000 customers at Ron’s grocery store showed that

 35% (3,150 customers) would buy Grainy-Os cereal

- In the first month, 4,200 out of the 9,000 customers bought the cereal

∴ The approximated value is 3150

∴ The exact value is 4200

∴ The percent of error = × 100%

∴ The percent of error = I-0.25I × 100% = 25%

* The percent error in the survey is 25%

●yes, this statement is absolutely apt.

hope this you

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.