Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class p
1 answer:
Answer:
a) p-hat (sampling distribution of sample proportions)
b) Symmetric
c) σ=0.058
d) Standard error
e) If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).
Step-by-step explanation:
a) This distribution is called the <em>sampling distribution of sample proportions</em> <em>(p-hat)</em>.
b) The shape of this distribution is expected to somewhat normal, symmetrical and centered around 16%.
This happens because the expected sample proportion is 0.16. Some samples will have a proportion over 0.16 and others below, but the most of them will be around the population mean. In other words, the sample proportions is a non-biased estimator of the population proportion.
c) The variability of this distribution, represented by the standard error, is:
d) The formal name is Standard error.
e) If we divided the variability of the distribution with sample size n=90 to the variability of the distribution with sample size n=40, we have:
If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).
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Answer:
(a) X ~ N().
~ N(
).
(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.
(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.
(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.
Step-by-step explanation:
We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.
Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.
(a) Let X = <u><em>amount of syrup that people put on their pancakes</em></u>
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = mean amount of syrup = 63 mL
= standard deviation = 13 mL
So, the distribution of X ~ N().
Let = <u><em>sample mean amount of syrup that people put on their pancakes</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = mean amount of syrup = 63 mL
= standard deviation = 13 mL
n = sample of people = 43
So, the distribution of ~ N(
).
(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)
P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X 61.4 mL)
P(X < 62.8 mL) = P( <
) = P(Z < -0.02) = 1 - P(Z
0.02)
= 1 - 0.50798 = 0.49202
P(X 61.4 mL) = P(
) = P(Z
-0.12) = 1 - P(Z < 0.12)
= 1 - 0.54776 = 0.45224
Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.
(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < < 62.8 mL)
P(61.4 mL < < 62.8 mL) = P(
< 62.8 mL) - P(
61.4 mL)
P( < 62.8 mL) = P(
<
) = P(Z < -0.10) = 1 - P(Z
0.10)
= 1 - 0.53983 = 0.46017
P(
61.4 mL) = P(
) = P(Z
-0.81) = 1 - P(Z < 0.81)
= 1 - 0.79103 = 0.20897
Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.
(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.