Tickets for the baseball games were $2.50 for general admission and 50 cents for kids. If there were six times as many general a
1 answer:
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Answer:
n=369
Step-by-step explanation:
1) Notation and definitions
number of the selected labourers opt for a new incentive scheme.
random sample taken
estimated proportion of the selected labourers opt for a new incentive scheme.
true population proportion of the selected labourers opt for a new incentive scheme.
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
2) Solution tot he problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and
. And the critical value would be given by:
The margin of error for the proportion interval is given by this formula:
(a)
And on this case we have that and we are interested in order to find the value of n, if we solve n from equation (a) we got:
(b)
And replacing into equation (b) the values from part a we got:
And rounded up we have that n=369
Answer:
a. 68% of the workers will earn between $47300 and $69700.
b. 2.5% of workers will earn above $89000
c. Approximately 0
Step-by-step explanation:
The standard normal distribution curve in the attached graph is used to solve this question.
a. The value $47300 is a standard deviation below the mean i.e. 58500-11200=47300. While $69700 is a standard deviation above the mean. I.e. 58500+12000=69700.
Between the first deviation below and above the mean, you have 34%+34%=68% of the salary earners within this range. So we have 68%of staffs earning within this range
b. The second standard deviation above the mean is $80900. i.e. 58500+11200+11200=$80900
We have 50%+13.5%+2.5%= 97.5% earning below $80900. Therefore, 100-97.5= 2.5% of the workers earn above this amount.
c. From the Standard Deviation Rule, the probability is only about (1 -0 .997) / 2 = 0.0015 that a normal value would be more than 3 standard deviations away from its mean in one direction or the other. The probability is only 0.0002 that a normal variable would be more than 3.5 standard deviations above its mean. Any more standard deviations than that, and we generally say the probability is approximately zero.
