Exam 2
Find the probability of the indicated event if P(E)equals 0.30 and P(F)equals0.45. P(E and F) = .10
.3 + .45 - .1 = .65
The graph of a normal curve is given. Use the graph to identify the value of mu and sigma.
u = 530 [middle value of graph] o = 15 [difference between values, (530-515)]
In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1,050 kWh and a standard deviation of 218 kWh. Find Upper P 45, which is the consumption level separating the bottom 45% from the top 55%. Round to the nearest tenth.
1,022.6
z_0.09
1.34
The table below shows the soft drinks preferences of people in three age groups. cola root beer lemon-lime under 21 years of age 40 25 20 between 21 and 40 35 20 30 over 40 years of age 20 30 35 If one of the 255 subjects is randomly selected, find the probability that the person is over 40 and drinks cola. Express your answer as a simplified fraction.
20/255 = 4/51
Nonsmoker Light smoker Heavy smoker Total Men 337 32 48 417 Women 315 49 49 413 Total 652 81 97 830 If two different people are randomly selected from the 830 subjects (without replacement), find the probability that they are both women.
413/830 * 412/829 = .2473
An investment counselor calls with a hot stock tip. He believes that if the economy remains strong, the investment will result in a profit of $50 comma 000. If the economy grows at a moderate pace, the investment will result in a profit of $10 comma 000. However, if the economy goes into recession, the investment will result in a loss of $50 comma 000. You contact an economist who believes there is a 20% probability the economy will remain strong, a 60% probability the economy will grow at a moderate pace, and a 20% probability the economy will slip into recession. What is the expected profit from this investment?
50000(.2)+10000(.6)-50000(.2) = 6000
About _____% of the area is between z = -2 and z = 2 (or within 2 standard deviations of the mean).
95.45%
Find the probability of the indicated event if P(E)=0.30 and P(F)=0.45 and P(F|E)=0.667
P(E and F) = 0.3*.667 = .200 P(E or F) = (.3+.45) - .200 = .55
The shape of the distribution of the time required to get an oil change at a 10-minute oil-change facility is unknown. However, records indicate that the mean time is 11.4 minutes, and the standard deviation is 4.7 minutes. Complete parts (a) and (b). (b) What is the probability that a random sample of n=40 oil changes results in a sample mean time less than 10 minutes?
The sample size needs to be greater than or equal to 30. z = (10-11.4)/4.7/sqrt(40) = -1.88 z < -1.88 = .0301
The distribution of the sample mean, x overbar, will be normally distributed if the sample is obtained from a population that is normally distributed, regardless of the sample size.
True
Assume that a study of 500 randomly selected school bus routes showed that 477 arrived on time. Is it "unusual" for a school bus to arrive late?
Yes
In a large state, 54% of adults have health insurance. If 4 adults are selected at random from this state, find the following probabilities. (Note: The population of the state is large enough that the choice of each adult can be considered independent) a. Find the probability that 4 adults selected at random all have health insurance. b. Find the probability that none of the 4 have health insurance. c. Find the probability that at least one of the 4 has health insurance.
a. (.54)^4 = .0850 b. 1-.54 = .46; .46^4 = .0448 c. 1-.0448 = .9552
The table below describes the smoking habits of a group of asthma sufferers. Nonsmoker Occasional smoker Regular smoker Heavy smoker Total Men 366----- 50 -------89 ----------44 --------549 Women 438----- 46 -------88 ----------33 --------605 Total 804----- 96 -------177 ----------77 --------1,154 a. If one person is randomly selected, find the probability that the person smokes. b. If one person is randomly selected, find the probability that the person is a man or a heavy smoker. c. If one person is randomly selected, find the probability that the person is a non-smoking woman. d. Find the probability that the chosen person is a man, given that the person is a heavy smoker. e. Given that the person is a man, find the probability he is an occasional smoker.
a. (96+177+77)/1154 = .303 b. 549/1154 + 77/1154 - 44/1154 = .504 c. 438/1154 = .380 d. 50/549 = .091
The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 7.8 cm. a. Find the probability that an individual distance is greater than 215.50 cm. b. Find the probability that the mean for 15 randomly selected distances is greater than 204.00 cm. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
a. .0999 b. .7718 (new std. dev. is 7.8/sqrt(204)) c. The normal distribution can be used because the original population has a normal distribution.
x P(x) 0 0.020 1 0.065 2 0.161 3 0.291 4 0.236 5 0.136 6 0.091 (a) Find he probability that an employee worked more than 4 hours of overtime. (b) Find the probability that an employee worked at most 2 hours of overtime. (c) Find the probability that an employee worked at least one hour of overtime. (d) Find the mean, variance, and standard deviation of the probability distribution. (e) Interpret the results in the context of the real-life situation.
a. .136 + .091 = .277 b. .02 + .065 + .161 = .246 c. .065 + .161 + .291 + .236 + .136 + .091 = .98 d. mean = sum(x*"P(x)") = 3.4 std. dev. = sqrt(mean/sqrt(7)) = 1.4 e. An employee works an average of approximately 3.4 overtime hours per week.
According to an annual crime report in a certain county, 54% of murders end in an arrest. Complete parts (a) through (d). (a) For 500 randomly selected murders, compute the mean and standard deviation of the random variable X, the number of murders end in an arrest. (b) Interpret the mean. Choose the correct answer below. (c) Of the 500 randomly selected murders, find the interval that would be considered "usual" for the number of murders that were cleared. Choose the correct answer below. (d) Would it be unusual if more than 418 of the 500 murders were cleared?
a. .54*500 = 270 sqrt(500*.54*.46) = 11.1 b. For every 500 murders, 270.0 of them would be expected to end in arrest, on average. c. 270-2(11.1) = 248 270+2(11.1) = 292 248 - 292 d. Yes
In a certain class of students, there are 11 boys & 9 girls from Wilmette, 2 boys & 5 girls from Kenilworth and 7 boys & 4 girls from Glencoe. (Creating a two-way table may help) a. If the teacher calls upon a student to answer a question, what is the probability that the student will be from Kenilworth? Round your answer to three decimal places. b. If the teacher calls upon a student to answer a question, what is the probability that the student will be a boy? Round your answer to three decimal places. c. If the teacher calls upon a student to answer a question, what is the probability that the student will be a boy OR will be from Kenilworth? Round your answer to three decimal places.
a. 7/38 = .184 b. 20/38 = .526 c. 7/38 + 20/38 - 2/38 = .658
According to an airline, a particular flight is on time 75% of the time. Suppose 20 flights are randomly selected and the number of on time flights is recorded. left parenthesis a right parenthesis Determine whether this is a binomial experiment. (b) Find the probability that exactly 18 flights are on time. (c) Find the probability that at least 18 flights are on time. (d) Find the probability that fewer than 18 flights are on time. (e) Find the probability that between 16 and 18 flights, inclusive, are on time.
a. Yes, because the experiment satisfies all the criteria for a binomial experiment. [binomial calc -> n=20, p=.75] b. p(x = 18) = .0669 c. p(x >= 18) = .0913 d. p(x < 18) = .9087 e. p(16 <= x <= 18) = .3905
Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 240 days and standard deviation sigma equals 14 days. Complete parts (a) through (f) below. (a) What is the probability that a randomly selected pregnancy lasts less than 235 days? (b) Suppose a random sample of 17 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. (c) What is the probability that a random sample of 17 pregnancies has a mean gestation period of 235 days or less?
a. [normal calc -> mean=240, std. dev. = 14] p(x <= 235) = .3605 If 100 pregnant individuals were selected independently from this population, we would expect 36 pregnancies to last less than 235 days. b. The sampling distribution of x-bar is normal with u_x-bar = 240 and o_x-bar = 14/sqrt(17) = 3.3955 c. [normal calc -> mean=240, std. dev. = 3.3955] The probability that the mean of a random sample of 17 pregnancies is less than 235 days is approximately 0.0704.
The lengths of pregnancies are normally distributed with a mean of 269 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 309 days or longer. b. If the length of pregnancy is in the lowest 2%, then the baby is premature. Find the length that separates premature babies from those who are not premature.
a. [normal calc -> mean=269, std. dev.=15] p(x >= 309) = .0038 b. [normal calc -> mean=269, std. dev.=15] p(x <= ____) = .02 -> _____ = 238
The reading speed of second grade students in a large city is approximately normal, with a mean of 92 words per minute (wpm) and a standard deviation of 10 wpm. a) What is the probability a randomly selected student in the city will read more than 98 words per minute? (b) What is the probability that a random sample of 10 second grade students from the city results in a mean reading rate of more than 98 words per minute? (c) There is a 5% chance that the mean reading speed of a random sample of 22 second grade students will exceed what value?
a. [normal calc -> mean=92, std. dev. = 10] p(x >= 98) = .2743 If 100 different students were chosen from this population, we would expect 27 to read more than 98 words per minute. b. new std. dev. 3.16227766 p(x >= 98) = .0289 c. new std. dev. 2.13200716 p(x >= ____) = .05, _____ is 95.51
In a past election, the voter turnout was 53%. In a survey, 1156 adults were asked if they voted in the election. a. Find the mean and standard deviation for the number of expected voters among any random group of 1156 adults. b. In the survey of 1156 adults, 631 said that they voted in the election. Is this result consistent with the turnout, or is this result unlikely to occur with a turnout of 53%? Why or why not?
a. mean = 1156*.53 = 612.7 std. dev. = sqrt(1156*.53*.47) = 17.0 b. This result is usual because 631 is within the range of usual values
Your wallet has 9 $10 bills and 4 $20 bills, and you reach in and pick TWO bills, without replacement.
highest: $40 lowest: $20 yes, the two bills could also add up to $30 dollars 20: 9/13 * 8/12 = .462 30: 9/13 * 4/12 = .231 * 2 = .462 40: 4/13 * 3/12 = .077
Find the Z-scores that separate the middle 11% of the distribution from the area in the tails of the standard normal distribution.
normal calc -> P(___ <= x <= ____) = .11 -.14, .14