pnorm(2, mean = 5, sd = 2)## [1] 0.06680728.1
15
x bar is approximately normal with mean = 80, standard deviation x = 2
P(x bar > 83) = 0.0668, if we take 100 simple random samples of size n = 49 from a population with mew = 80 and standard deviation = 14, then about 7 of the samples will result in a mean that is greater than 83.
P(x bar is less than or equal to 75.8) = 0.0179, if we take 100 simple random sample size of n = 49 from a population with mew = 80 and standard deviation = 14, then about 2 of the samples will result in a mean that is less than or equal to 75.8.
P(78.3 < x bar < 85.1) = 0.7969, if we take 100 simple random samples of size n = 49 from a population with mew = 80 and standard deviation = 14, then about 80 ofthe samples will result in a mean that is between 78.3 and 85.1.
17
The population must be normally distributed to compute probabilities involving the sample mean. If the population is normally distributed, then the sampling distribution of x bar is also normally distributed with mew = 64 and sigma x bar = 17 divided by the square root of 12 which equals approximately 4.907
P(x bar < 67.3) = 0.7486, if we take 100 simple random sample of size n=12 from a population that is normally distributed with mew = 64 and standard deviation = 17, then about 75 of the samples will result in a mean that is less than 67.3
P(x bar is greater than or equal to 65.2) = 0.4052, if we take 100 simple random samples of size n=12 from a population that is normally distributed with mew = 64 and standard deviation = 17,then about 40 or 41 of the samples will result in a mean that is greater than or equal to 65.2
19
P(X<260) = 0.3520, if we randomly select 100 human pregnancies, then about 35 of the pregnancies will last less than 260 days.
The sampling distribution of x bar is normal with mew = 266 and standard deviation = 16/square root of 20 which equals approximately 3.578
P(sample x is less than or equal to 260) = 0.0465, if we take 100 simple random samples of size n = 20 human pregancies, then about 5 of the samples will result in a mean gestation period of 260 days or less.
P(x bar is less than or equal to 260) = 0.0040, if we take 1000 simple random samples of size n = 50 human pregancies, then about 4 of the samples will result in a mean gestation period of 260 days or less.
This result would be unusual, so the sample likely came from a population whose mean gestation period is less than 266 days.
P(256 is less than or equal to x bar which is less than or equal to 276) = 0.9844, if we take 100 simple random samples of size n = 15 human pregnancies, then about 98 of the samples will result in a mean gestation period between 256 and 276 days, inclusive.
21
P(X > 95) = 0.3085, if we select a simple random sample of n = 100 second grade students, then about 31 of the students will read more than 95 words per minute.
P(x bar > 95) = 0.0418, if we take 100 simple random samples of size n = 12 second grade students, then about 4 of the samples will result in a mean reading rate that is more than 95 words per minute.
P(x bar > 95) = 0.0071, if we take 1000 simple random samples of size n = 24 second grade students, then about 7 of the samples will result in a mean reading rate that is more than 95 words per minute.
Increasing the sample size decreaes P(x bar > 95). This happens because standard deviation decreases as n increases.
A mean rating of 92.8 wpm is not unusual since P(x bar is greater than or equal to 92.8) = 0.1056. This means that the new reading program is not abundantly more effective than the old program.
There is a 5% chance that the mean reading speed of a random sample of 20 second grade students will exceed 93.7 words per minute.
23
P(X > 0) = 0.5675, if we select a simple random sample of n = 100 months, then about 57 of the months will have positive rates of return.
P(x bar > 0) = 0.7291, if we take 100 simple random samples of size n = 12 months, then about 73 of the samples will result in a mean monthly rate of return that is positive.
P(x bar > 0) = 0.8051, if we take 100 simple random samples of size n = 24 months, then about 81 of the samples will result in a mean monthly rate of return that is positive.
P(x bar > 0) = 0.8531, if we take 100 simple random samples of size n = 36 months, then about 85 of the samples will result in a mean monthly rate of return that is positive.
The likelihood of earning a positive rate of return increases as the investment time horizon increases.
8.2
11
The sampling distribution of p hat is approximately normal with the standard deviation = approximately 0.046 and mean = 0.8
P(p hat is greater than or equal to .84) = 0.1922. About 19 out of 100 random samples of size n=75 will result in 63 or more individuals with the characteristic.
P(p hat is less than or equal to 0.68) = 0.0047. About 5 out of 1000 random samples of size n = 75 will result in 51 or fewer indiuviduals with the characteristic, that is 68% or less.
12
The sampling distribution is approximately normal with a standard deviation of 0.337 and a mean of 0.65
P( p hat is greater than or equal to 0.68) = .1867
P( p hat is less than or equal to .59) = .8577
13
The sampling distribution of p hat is approximately normal with mean = 0.35, standard deviation approximately 0.015
P( p hat is greater than or equal to 0.39) = 0.0040. About 4 out of 1000 random samples of size n=1000 will result in 390 or more individuals with the characteristics, it’s 39% or more.
P( p hat is less than or equal to 0.32) = 0.0233. About 2 out of 100 random samplse of size n=1000 will result in 320 or fewer indivuiduals with the characterisitc, that is 32% or less.
14
The distribution is approximately a normal distribution. Mean = .42, standard deviation = .1292
P(x is greater than or equal to .657) = .0301
P(x is less than or equal to 584) = .9082
15
Qualitative with 2 possible outcomes - order a meal in a foreign language or not.
The source of the variabilility is the individuals in the survey and their ability to order a meal in a foreign language.
The sampling distribution of p hat is approximately normal with mean = 0.47 and standard deviation = about 0.035.
P(p hat > 0.5) = 0.1977. About 2 out of 100 random samples of size n = 200 Americans will result in more than 100 indivuiduals who can order a meal in a foreign language, more than 50%.
P(p hat less than or equal to 0.4) = 0.0239. About 2 out of 100 random samples of size n=200 Americans will result in more 80 or fewer individuals who can order a meal in a foreign language, that is 40% or less.
16
Quantitative with mulitple outcomes, people could have various levels of how satisfied with how their lives are going. It is not just a “yes” or “no” answer.
Because the value of the sample proportion can vary from sample to sample, it’s a radnom variable and has a probability distribution. If you do 2 surveys with 2 different groups of people, you’ll get different answers.
It’s normal because the smaple size (100) is smaller than the populatin size of all Americans and np(1-p) is 14.76 which is greater than 10. The new mean is .82 and the new standard deviation is 0.38
P(x>.85) = .4721
It would be unusual because 75 is too big.
17
The sampling distribution of p hat is approximately normal with mean = 0.39 and standard deviation = about 0.022.
P(p hat < 0.38) = 0.3228. About 32 out of 100 random samples of size n = 500 adult Americans will result in fewer than 190 individuals who believe that marriage is obsolete, less than 38%
P(0.40 < p hat < 0.45) = 0.3198. About 32 out of 100 random samples of size n = 500 adult Americans will result in between 200 and 225 indivudials who beliebe that marriage is obsolete, between 40% - 45%.
P(p hat is greater than or equal to 0.42) = 0.0838. About 8 out of 100 random samples of size n=500 adult Americans will result in 210 or more indivudals who beliebe that marriage is obsolete. This result is not unusual, this is 42% or more.