# Assume that the random variable Z follows standard normal distribution, calculate the following probabilities (Round to two decimal places) a)P(z>1.9) b)P(−2<=z<=1.2) c)P(z>−0.2)

Question
Normal distributions
Assume that the random variable Z follows standard normal distribution, calculate the following probabilities (Round to two decimal places)
a)P(z>1.9)
b)$$\displaystyle{P}{\left(−{2}\le{z}\le{1.2}\right)}$$
c)P(z>−0.2)

2020-11-06
Given:
random variable z follows the standard normal distributions.
a)$$\displaystyle{P}{\left({z}{>}{1.9}\right)}={1}−{P}{\left({z}\le{1.9}\right)}$$
P(z>1.9) = 1−0.9713
$$\displaystyle{P}{\left({z}{>}{1.9}\right)}={0.0287}\approx{0.03}$$
b)$$\displaystyle{P}{\left(−{2}\le{z}\le{1.2}\right)}={P}{\left({z}\le{1.2}\right)}−{P}{\left({z}{<}−{2}\right)}$$</span>
$$\displaystyle{P}{\left(−{2}\le{z}\le{1.2}\right)}={0.8849}−{0.0228}$$
$$\displaystyle{P}{\left(−{2}\le{z}\le{1.2}\right)}={0.8621}\approx{0.86}$$
c)$$\displaystyle{P}{\left({z}{>}−{0.2}\right)}={1}−{P}{\left({z}\le−{0.2}\right)}$$
P(z>−0.2) = 1−0.4207
$$\displaystyle{P}{\left({z}{>}−{0.2}\right)}={0.5793}\approx{0.58}$$

### Relevant Questions

A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age.
$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old.
$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
(i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$
(a) What is the level of significance?
State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
Would you rather spend more federal taxes on art? Of a random sample of $$n_{1} = 86$$ politically conservative voters, $$r_{1} = 18$$ responded yes. Another random sample of $$n_{2} = 85$$ politically moderate voters showed that $$r_{2} = 21$$ responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use $$\alpha = 0.05.$$ (a) State the null and alternate hypotheses. $$H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2$$
$$H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}$$ (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference $$p_{1} - p_{2}$$. Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are not statistically significant. (f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{H}{o}{u}{s}{e}{w}{\quad\text{or}\quad}{k}{H}{o}{u}{r}{s}\backslash{h}{l}\in{e}{G}{e}{n}{d}{e}{r}&{S}{a}\mp\le\ {S}{i}{z}{e}&{M}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {D}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{W}{o}{m}{e}{n}&{473473}&{33.133}{.1}&{14.214}{.2}\backslash{h}{l}\in{e}{M}{e}{n}&{488488}&{18.618}{.6}&{15.715}{.7}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ a. Based on this​ study, calculate how many more hours per​ week, on the​ average, women spend on housework than men. b. Find the standard error for comparing the means. What factor causes the standard error to be small compared to the sample standard deviations for the two​ groups? The cause the standard error to be small compared to the sample standard deviations for the two groups. c. Calculate the​ 95% confidence interval comparing the population means for women Interpret the result including the relevance of 0 being within the interval or not. The​ 95% confidence interval for ​$$\displaystyle{\left(\mu_{{W}}-\mu_{{M}}​\right)}$$ is: (Round to two decimal places as​ needed.) The values in the​ 95% confidence interval are less than 0, are greater than 0, include 0, which implies that the population mean for women could be the same as is less than is greater than the population mean for men. d. State the assumptions upon which the interval in part c is based. Upon which assumptions below is the interval​ based? Select all that apply. A.The standard deviations of the two populations are approximately equal. B.The population distribution for each group is approximately normal. C.The samples from the two groups are independent. D.The samples from the two groups are random.
a) Which of the following properties distinguishes the standard normal distribution from other normal distributions?
-The mean is located at the center of the distribution.
-The total area under the curve is equal to 1.00.
-The curve is continuous.
-The mean is 0 and the standard deviation is 1.
b) Find the probability $$\displaystyle{P}{\left({z}{<}-{0.51}\right)}$$ using the standard normal distribution.
c) Find the probability $$\displaystyle{P}{\left({z}{>}-{0.59}\right)}$$ using the standard normal distribution.
factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of examination scores is very small, it usually means that the examination was either too hard or too easy. However, if the variance of scores is moderately large, then there is a definite difference in scores between "better," "average," and "poorer" students. A group of attorneys in a Midwest state has been given the task of making up this year's bar examination for the state. The examination has 500 total possible points, and from the history of past examinations, it is known that a standard deviation of around 60 points is desirable. Of course, too large or too small a standard deviation is not good. The attorneys want to test their examination to see how good it is. A preliminary version of the examination (with slight modifications to protect the integrity of the real examination) is given to a random sample of 20 newly graduated law students. Their scores give a sample standard deviation of 70 points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new examination is 60 against the claim that the population standard deviation is different from 60.
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H_{0}:\sigma=60,\ H_{1}:\sigma\ <\ 60H_{0}:\sigma\ >\ 60,\ H_{1}:\sigma=60H_{0}:\sigma=60,\ H_{1}:\sigma\ >\ 60H_{0}:\sigma=60,\ H_{1}:\sigma\ \neq\ 60$$
(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original distribution?
We assume a binomial population distribution.We assume a exponential population distribution. We assume a normal population distribution.We assume a uniform population distribution.
Is statistical inference intuitive to babies? In other words, are babies able to generalize from sample to population? In this study,1 8-month-old infants watched someone draw a sample of five balls from an opaque box. Each sample consisted of four balls of one color (red or white) and one ball of the other color. After observing the sample, the side of the box was lifted so the infants could see all of the balls inside (the population). Some boxes had an “expected” population, with balls in the same color proportions as the sample, while other boxes had an “unexpected” population, with balls in the opposite color proportion from the sample. Babies looked at the unexpected populations for an average of 9.9 seconds (sd = 4.5 seconds) and the expected populations for an average of 7.5 seconds (sd = 4.2 seconds). The sample size in each group was 20, and you may assume the data in each group are reasonably normally distributed. Is this convincing evidence that babies look longer at the unexpected population, suggesting that they make inferences about the population from the sample? Let group 1 and group 2 be the time spent looking at the unexpected and expected populations, respectively. A) Calculate the relevant sample statistic. Enter the exact answer. Sample statistic: _____ B) Calculate the t-statistic. Round your answer to two decimal places. t-statistic = ___________ C) Find the p-value. Round your answer to three decimal places. p-value =
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}$$
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
(b) $$Pr(z>2.5)$$
Consider two independent populations that are normally distributions. A simple random sample of $$\displaystyle{n}_{{1}}={41}$$ from the first population showed $$\displaystyle\overline{{x}}_{{11}}={33}$$, and a simple random of size $$\displaystyle{n}_{{2}}={48}$$ from the second population showed
$$\displaystyle\overline{{x}}_{{2}}={32}$$
Suppose $$\displaystyle{s}_{{1}}={9}{s}_{{1}}={9}{\quad\text{and}\quad}{s}_{{2}}={10}{s}_{{2}}={10}$$, find a 98% confidence interval for $$\displaystyle\mu_{{1}}−\mu_{{2}}\mu_{{1}}-\mu_{{2}}$$. (Round answers to two decimal places.)
(a) $$Pr(0 \leq Z \leq 2.5)$$