# The score of adults on an IQ test are approximately Normal with mean 100 and standart deviation 15. Alysha scores 135 on such a test. She scores highter than what percent of all adults? a. About 5% b. About 95% c. About 99%

Question
Probability
The score of adults on an IQ test are approximately Normal with mean 100 and standart deviation 15. Alysha scores 135 on such a test. She scores highter than what percent of all adults?

2020-11-13
Let X be the random variable of IQ test. Then $$\displaystyle{X}\sim{N}{\left({100},{15}^{{2}}\right)}$$
which means that $$\displaystyle\frac{{{X}-{100}}}{{15}}\sim{N}{\left({0},{1}\right)}$$
We need to find $$\displaystyle{P}{\left({X}\le{135}\right)}$$
We write this is a way to get the standard normal variable: $$\displaystyle{P}{\left({\left(\frac{{{X}-{100}}}{{15}}\right)}\le\frac{{{135}-{100}}}{{15}}\right)}={P}{\left({\left(\frac{{{X}-{100}}}{{15}}\right)}\le\frac{{7}}{{3}}\right)}$$
Let Ф denote the standard normal distribution, so $$\displaystyle{P}{\left({\left(\frac{{{X}-{100}}}{{15}}\right)}\le\frac{{7}}{{3}}\right)}=Ф{\left(\frac{{7}}{{3}}\right)}\simФ{\left({2.33}\right)}\sim{0.9901}$$
Thus, (c) is correct.

### Relevant Questions

1. Find each of the requested values for a population with a mean of $$? = 40$$, and a standard deviation of $$? = 8$$ A. What is the z-score corresponding to $$X = 52?$$ B. What is the X value corresponding to $$z = - 0.50?$$ C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of $$M=42$$ for a sample of $$n = 4$$ scores? E. What is the z-scores corresponding to a sample mean of $$M= 42$$ for a sample of $$n = 6$$ scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: $$a. -2.00 b. 1.25 c. 3.50 d. -0.34$$ 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with $$\mu = 78$$ and $$\sigma = 12$$. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: $$82, 74, 62, 68, 79, 94, 90, 81, 80$$. 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $$12 (\mu = 42, \sigma = 12)$$. You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is$44.50 from tips. Test for a difference between this value and the population mean at the $$\alpha = 0.05$$ level of significance.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{H}{o}{u}{s}{e}{w}{\quad\text{or}\quad}{k}{H}{o}{u}{r}{s}\backslash{h}{l}\in{e}{G}{e}{n}{d}{e}{r}&{S}{a}\mp\le\ {S}{i}{z}{e}&{M}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {D}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{W}{o}{m}{e}{n}&{473473}&{33.133}{.1}&{14.214}{.2}\backslash{h}{l}\in{e}{M}{e}{n}&{488488}&{18.618}{.6}&{15.715}{.7}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ 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.
For a variable that is normally distributed with a mean of 50 and a standard deviation of 5, approximately _? of the scores are between _?.
A)95%, 45 and 55
B)95%, 40 and 60
C)68%, 40 and 60
D)95%, 35 and 65
E)99%, 40 and 60
1. A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 44.7 texts per day. The standard deviation was 16.5 texts. a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean number of texts per day is between and texts. c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day. 2. You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately $$\displaystyle\sigma={40.4}$$ dollars. You would like to be 90% confident that your estimate is within 1.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? n = 3. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $$\displaystyle\sigma={57.5}$$. You would like to be 95% confident that your estimate is within 0.1 of the true population mean. How large of a sample size is required?
a. If the data is weight, the z-score for someone who is overweight would be
-positive
-negative
-zero
b. If the data is IQ test scores, an individual with a negative z-score would have a
-high IQ
-low IQ
-average IQ
c. If the data is time spent watching TV, an individual with a z-score of zero would
-watch very little TV
-watch a lot of TV
-watch the average amount of TV
d. If the data is annual salary in the U.S and the population is all legally employed people in the U.S., the z-scores of people who make minimum wage would be
-positive
-negative
-zero
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.
The distribution of height for a certain population of women is approximately normal with mean 65 inches and standard deviation 3.5 inches. Consider two different random samples taken from the population, one of size 5 and one of size 85.
Which of the following is true about the sampling distributions of the sample mean for the two sample sizes?
Both distributions are approximately normal with mean 65 and standard deviation 3.5.
A
Both distributions are approximately normal. The mean and standard deviation for size 5 are both less than the mean and standard deviation for size 85.
B
Both distributions are approximately normal with the same mean. The standard deviation for size 5 is greater than that for size 85.
C
Only the distribution for size 85 is approximately normal. Both distributions have mean 65 and standard deviation 3.5.
D
Only the distribution for size 85 is approximately normal. The mean and standard deviation for size 5 are both less than the mean and standard deviation for size 85.
E
American automobiles produced in 2012 and classified as “large” had a mean fuel economy of 19.6 miles per gallon with a standard deviation of 3.36 miles per gallon. A particular model on this list was rated at 23 miles per gallon, giving it a z-score of about 1.01. Which statement is true based on this information? A) Because the standard deviation is small compared to the mean, a Normal model is appropriate and we can say that about 84.4% of “large” automobiles have a fuel economy of 23 miles per gallon or less. B) Because a z-score was calculated, it is appropriate to use a Normal model to say that about 84.4% of “large” automobiles have a fuel economy of 23 miles per gallon or less. C) Because 23 miles per gallon is greater than the mean of 19.6 miles per gallon, the distribution is skewed to the right. This means the z-score cannot be used to calculate a proportion.D) Because no information was given about the shape of the distribution, it is not appropriate to use the z-score to calculate the proportion of automobiles with a fuel economy of 23 miles per gallon or less. E) Because no information was given about the shape of the distribution, it is not appropriate to calculate a z-score, so the z-score has no meaning in this situation.
Montarello and Martins (2005) found that fifth grade students completed more mathematics problems correctly when simple problems were mixed in with their regular math assignments. To further explore this phenomenon, suppose that a researcher selects a standardized mathematics achievement test that produces a normal distribution of scores with a mean of $$\mu= 100\ and\ a\ standard\ deviation\ of\ \sigma = 24$$. The researcher modifies the test by inserting a set of very easy problems among the standardized questions and gives the modified test to a sample of n = 36 students. If the average test score for the sample is M = 120, is this result sufficient to conclude that inserting the easy questions improves student performance? Use a one-tailed test with $$\alpha = .05$$.
$$H_{0}:\sigma=60,\ H_{1}:\sigma\ <\ 60H_{0}:\sigma\ >\ 60,\ H_{1}:\sigma=60H_{0}:\sigma=60,\ H_{1}:\sigma\ >\ 60H_{0}:\sigma=60,\ H_{1}:\sigma\ \neq\ 60$$