Stats215 Ch 6

Assume that adults have IQ scores that are normally distributed with a mean of μ=100 and a standard deviation σ=20. Find the probability that a randomly selected adult has an IQ between 86 and 114.

0.516 find the area of both (use the Z score formula) subtract smallest number from the largest

3 conclusions of the Central Limit Theorem

1) As the sample size increases (n > 30), the distribution of the sample means starts approaching a normal distribution no matter what the original (population) distribution was. 2) The mean of the sample means is equal to the population mean. (μ (of xbar) = μ) 3) The standard deviation of the sample means is equal to the population standard deviation divided by the square root of the sample size n. (σ(of xbar) = σ / √n) Also called the standard error of the mean. Therefore the formula for the z-score (standard score) for a sample mean is z = [x (bar) - μ(of xbar)] / σ(of xbar) = (x(bar) - μ) / (σ / √n) = [(xbar - μ) / 1] X (√n / σ) = [(xbar - μ) / σ ] X √n

Requirements of a density Curve

1. The curve cannot fall below the horizontal axis. 2. Every point on the curve must have a vertical height that is 0 or greater. 3. The total area under the curve must equal 1.

normal distribution of a random​ variable described by

1. The graph of the distribution is​ bell-shaped. 2. The graph is centered around the mean. 3. The graph of the distribution is symmetric.

standard normal​ distribution factors

1. The total area under the curve must equal 1. 2. The graph is symmetric. 3. It is a normal distribution with a mean of 0 and a standard deviation of 1.

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 9 minutes. Find the probability that a randomly selected passenger has a waiting time less than 3.75 minutes.

9 possible minutes, area must be equal to 1 1/9 is the height of the graph. so 3.75 (width) times 1/9 = 0.4166666

following are true about Z scores

A.If values are converted to standard​ z-scores, then procedures for working with all normal distributions are the same as those for the standard normal distribution. B.A​ z-score is a conversion that standardizes any value from a normal distribution to a standard normal distribution. C.The area in any normal distribution bounded by some score x is the same as the area bounded by the equivalent​ z-score in the standard normal distribution.

Which of the following is not​ true?

A​ z-score is not an area under the normal curve. Distances along the horizontal axis are represented by​ z-scores, while regions under the curve are represented by areas.

Assume that thermometer readings are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A thermometer is randomly selected and tested. For the case​ below, draw a​ sketch, and find the probability of the reading.​ (The given values are in Celsius​ degrees.) Between 0.50 and 2.25

Draw a sketch. Choose the correct graph below. see what the histogram is asking and area needed - the sketch must have a mean of 0 and sd of 1 The probability of getting a reading between 0.50°C and 2.25°C is 0.2963 the high area subtracted from the low area, so in this case from the table it is 0.9878 - 0.6915 ​(Round to four decimal places as​ needed.)

Three randomly selected children are surveyed. The ages of the children are 1​, 4​, and 10. Assume that samples of size n=2 are randomly selected with replacement from the population of 1​, 4​, and 10. Listed below are the nine different samples. Complete parts​ (a) through​ (d).

Find the value of the population variance σ ^2. 14 (Use the list and 1variabl on TI83) Find the mean of the sampling distribution of the sample variance. 14 -> add all the variances together divide by 9 Based on the preceding​ results, is the sample variance an unbiased estimator of the population​ variance? Why or why​ not? The sample variance targets the population variance. As​ such, the sample variance is an unbiased estimator of the population variance.​​

Which of the following is not a commonly used​ practice

If the distribution of the sample means is normally​ distributed, and n>​30,then the population distribution is normally distributed. (This is not always the case because skewed population distributions may have sampling distributions that are normally distributed.)

Weights of golden retriever dogs are normally distributed. Samples of weights of golden retriever​ dogs, each of size n=​15, are randomly collected and the sample means are found. Is it correct to conclude that the sample means cannot be treated as being from a normal distribution because the sample size is too​ small? Explain.

No; the original population is normally​ distributed, so the sample means will be normally distributed for any sample size.

What's wrong with the following​ statement? ​"Because the digits​ 0, 1,​ 2, . . .​ , 9 are the normal results from lottery​ drawings, such randomly selected numbers have a normal​ distribution."

Since the probability of each digit being selected is​ equal, lottery digits have a uniform​ distribution, not a normal distribution.

states that​ if, under a given​ assumption, the probability of a particular observed event is exceptionally small​ (such as less than​ 0.05), we conclude that the assumption is probably not correct.

The Rare Event Rule for Inferential Statistics

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find the bone density test scores that can be used as cutoff values separating the lowest 19​% and highest 19​%, indicating levels that are too low or too​ high, respectively.

The bone density scores are −0.87, 0.87. on the chart, find the 0.19 area on the positive side, and do the area before but ensure 0.9999 - 0.19 is found and use 0.8099 which is 0.8079 at the z score 0.87, this is finding the score above on negative side find 0.19 which is closest to 0.1922 at z score -0.87 this is finding the z score below

Which of the following is NOT a property of the sampling distribution of the​ variance?

The distribution of sample variances tends to be a normal distribution. typically it will be skewed right

Which of the following is NOT a conclusion of the Central Limit​ Theorem?

The distribution of the sample data will approach a normal distribution as the sample size increases.

Which of the following is NOT a property of the sampling distribution of the sample​ mean?

The distribution of the sample mean tends to be skewed to the right or left. (normally sample mean is uniform following normal distribution)

What requirements are necessary for a normal probability distribution to be a standard normal probability​ distribution?

The mean and standard deviation have the values of μ=0 and σ=1.

Which of the following is a biased​ estimator? That​ is, which of the following does not target the population​ parameter?

The median is a biased estimator.

Assume that human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.64°F. a. A hospital uses 100.6°F as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a​ fever? Does this percentage suggest that a cutoff of 100.6°F is​ appropriate? b. Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature​ be, if we want only​ 5.0% of healthy people to exceed​ it? (Such a result is a false​ positive, meaning that the test result is​ positive, but the subject is not really​ sick.

The percentage of normal and healthy persons considered to have a fever is 0.01​%. Find area of Z, subtract it from 1 to find area to the right. Yes this is appropriate, because there is a small probability that a normal and healthy person would be considered to have a fever. b. The minimum temperature for requiring further medical tests should be 99.25°F if we want only​ 5.0% of healthy people to exceed it. to find the Z area, it is 95% (the area we want below X temp)

A survey found that​ women's heights are normally distributed with mean 63.9 in and standard deviation 2.4 in. A branch of the military requires​ women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest​ 1% and the tallest​ 2%, what are the new height​ requirements?

The percentage of women who meet the height requirement is 99.3​% Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall? no, because only a small percentage of women are not allowed to join this branch of the military because of their height. For the new height​ requirements, this branch of the military requires​ women's heights to be at least 58.3in and at most 68.8in. X is the new heights, use the formula to find both X = μ + σz

Which of the following is NOT a requirement for using the normal distribution as an approximation to the binomial​ distribution?

The sample is the result of conducting several dependent trials of an experiment in which the probability of success is p.

​_____________ is the distribution of all values of the statistic when all possible samples of the same size n are taken from the same population.

The sampling distribution of a statistic

​_____________ is the distribution of sample​ proportions, with all samples having the same sample size n taken from the same population.

The sampling distribution of the proportion (The sampling distribution of a statistic is typically represented as a probability distribution in the format of a​ table, probability​ histogram, or​ formula.)

The standard normal distribution is a probability distribution with these properties:

The standard normal distribution is a probability distribution with these properties: • The distribution is a normal distribution, so it is bell-shaped as in Figure 6-4. • The population parameter of the mean has the specific value of m=0. • The population parameter of the standard deviation has the specific value of s=1. • The total area under its density curve is equal to 1

A continuous random variable has a​ _______ distribution if its values are spread evenly over the range of possibilities.

UNIFORM

Which of the following would be information in a question asking you to find the area of a region under the standard normal curve as a​ solution?

When given a​ z-score, you are usually finding the area of the shaded region under the standard normal curve. For the standard normal​ curve, a​ z-score is a distance along the horizontal axis.

Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find P18​, the 18th percentile. This is the bone density score separating the bottom 18% from the top 82%

Which graph represents P18​? Choose the correct graph below. xP18 A symmetric bell-shaped curve is plotted over a horizontal x-scale. A vertical line segment runs from the scale to the curve at labeled coordinate P subscript 18, which is to the left of the curve's center and peak; the segment's height is approximately 65 percent of the height of the curve at its peak. The area under the curve to the left of the vertical line is shaded. Part 2 The bone density score corresponding to P18 is -0.92 (find the first negative that is in the 0.18xx don't use the 017xx) ​(Round to two decimal places as​ needed.)

If you are asked to find the 85th​ percentile, you are being asked to find​ _____.

a data value associated with an area of 0.85 to its left

The overhead reach distances of adult females are normally distributed with a mean of 202.5 cm and a standard deviation of 7.8 cm.

a. Find the probability that an individual distance is greater than 211.80cm. probability greater than 211.80 = 0.117 b. Find the probability that the mean for 15 randomly selected distances is greater than 200.3cm The probability is 0.8621 C. choose the correct answer The normal distribution can be used because the original population has a normal distribution.

Three randomly selected children are surveyed. The ages of the children are 2​, 4​, and 12. Assume that samples of size n=2 are randomly selected with replacement from the population of 2​, 4​, and 12. Listed below are the nine different samples. Complete parts​ (a) through​ (d).

a. Find the value of the population median. 4 b. Find the median of each of the nine​ samples, then summarize the sampling distribution of the medians in the format of a table representing the probability distribution of the distinct median values. Sample Median Probability 2 1/9 3 2/9 4 1/9 7 2/9 8 2/9 12 1/9 c. Find the mean of the sampling distribution of the sample median. 6 d. Based on the preceding​ results, is the sample median an unbiased estimator of the population​ median? Why or why​ not? The sample median does not target the population median. As​ such, the sample median does not make a good estimator of the population median

ssume a population of 1​, 3​, and 8. Assume that samples of size n=2 are randomly selected with replacement from the population. Listed below are the nine different samples. Complete parts a through d below

a. Find the value of the population standard deviation σ. σ=2.944 b. Find the standard deviation of each of the nine​ samples, then summarize the sampling distribution of the standard deviations in the format of a table representing the probability distribution of the distinct standard deviation values. Use ascending order of the sample standard deviations. s Probability 0 1/3 1.414 2/9 3.536 2/9 4.95 2/9 c. Find the mean of the sampling distribution of the sample standard deviations. The mean of the sampling distribution of the sample standard deviations is 2.22 d. Do the sample standard deviations target the value of the population standard​ deviation? In​ general, do sample standard deviations make good estimators of population standard​ deviations? Why or why​ not? The sample standard deviations do not target the population standard​ deviation, therefore, sample standard deviations are biased estimators.

hree randomly selected children are surveyed. The ages of the children are 1​, 2​, and 12. Assume that samples of size n=2 are randomly selected with replacement from the population of 1​, 2​, and 12.

a. For the​ population, find the proportion of odd numbers. The proportion is 0.333 b. Find the proportion of odd numbers of each of the nine​ samples, then summarize the sampling distribution of the sample proportion of odd numbers in the format of a table representing the probability distribution of the distinct proportion values. Sample Proportion Probability 0 4/9 0.5 4/9 1 1/9 ​ c. Find the mean of the sampling distribution of the sample proportion of odd numbers. The mean is 0.333 d. Based on the preceding​ results, is the sample proportion an unbiased estimator of the population ​ proportion? Why or why​ not? The sample proportions target the proportion of odd numbers in the​ population, so sample proportions make good estimators of the population proportion.

Assume that females have pulse rates that are normally distributed with a mean of μ=74.0 beats per minute and a standard deviation of σ=12.5 beats per minute. Complete parts​ (a) through​ (c) below.

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 80 beats per minute. The probability is 0.6844 >>> P(z), z= (80-74)/12.5 =0.48 ... P(0.48) = 0.6844 b. If 16 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 80 beats per minute. The probability is 0.9726 since we found that the zscore is 0.48 we can multiply by square root n (16) which is 4, so the new zscore is 1.92 with a probability of 0.9726 c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30? ---Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

When women were finally allowed to become pilots of fighter​ jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. The ejection seats were designed for men weighing between 120lb and 171lb. Weights of women are now normally distributed with a mean of 163 lband a standard deviation of 44 lb. Complete parts​ (a) through​ (c) below.

a. If 1 woman is randomly​ selected, find the probability that her weight is between 120 lb and 171 lb. The probability is approximately 0.4079 --- take P(171) - P(120) and that is the probability of the in between b. If 33 different women are randomly​ selected, find the probability that their mean weight is between 120lb and 171lb. 0.8507 multiply the z scores by the square root of 33 then subtract the probabilities to find the in between c. when redesigning the seats what is more important? The part​ (a) probability is more relevant because the seat performance for a single pilot is more important.

An airliner carries 200 passengers and has doors with a height of 78 in. Heights of men are normally distributed with a mean of 69.0 in and a standard deviation of 2.8 in. Complete parts​ (a) through​ (d).

a. If a male passenger is randomly​ selected, find the probability that he can fit through the doorway without bending. use 78-69 / 2.8 = 3.214 find the P(X) with the Z = 3.214 The probability is 0.9993 b. If half of the 200 passengers are​ men, find the probability that the mean height of the 100 men is less than 78 in. use the CTL for this = Z = ((xbar - mu) / σ) X √n The probability is [(78 - 69)/2.8]x √100 = 32.14 so the probability is going to be 0.9999 c. When considering the comfort and safety of​ passengers, which result is more​ relevant: the probability from part​ (a) or the probability from part​ (b)? Why? The probability from part​ (a) is more relevant because it shows the proportion of male passengers that will not need to bend D. When considering the comfort and safety of​ passengers, why are women ignored in this​ case?. Since men are generally taller than​ women, a design that accommodates a suitable proportion of men will necessarily accommodate a greater proportion of women.

For bone density scores that are normally distributed with a mean of 0 and a standard deviation of​ 1, find the percentage of scores that are

a. The percentage of bone density scores that are significantly high is 2.28​%. (1 minus the area from 2 is 0.0228 because it is everything above the sd of 2) b. The percentage of bone density scores that are significantly low is 2.28​%. c. The percentage of bone density scores that are not significant is 95.44​%. (everything between the sd 2 minus the -2 which is = 0.9772 - 0.0228 = 0.95544)

A​ gender-selection technique is designed to increase the likelihood that a baby will be a girl. In the results of the​ gender-selection technique, 854 births consisted of 448 baby girls and 406 baby boys. In analyzing these​ results, assume that boys and girls are equally likely.

a. The probability of getting exactly 448 girls in 854 births is 0.0097 ​(Round to four decimal places as​ needed.) Part 2 b. The probability of getting 448 or more girls in 854 births is 0.0803 ​(Round to four decimal places as​ needed.) Part 3 If boys and girls are equally​ likely, is 448 girls in 854 births unusually​ high? ​No, because 448 girls in 854 births is not far from what is​ expected, given the probability of having a girl or a boy. c. Which probability is relevant for trying to determine whether the technique is​ effective, the result from part​ (a) or the result from part​ (b)? The result from part​ (b) is more​ relevant, because one wants the probability of a result that is at least as extreme as the one obtained. d. Based on the​ results, does it appear that the​ gender-selection technique is​ effective? No​, because the probability of having 448 or more girls in 854 births is not ​unlikely, and​ thus, is attributable to random chance

An elevator has a placard stating that the maximum capacity is 3900 lb—26 passengers.​ So, 26 adult male passengers can have a mean weight of up to 3900/26=150 pounds. Assume that weights of males are normally distributed with a mean of 183 lb and a standard deviation of 29 lb. a. Find the probability that 1 randomly selected adult male has a weight greater than 150 lb. b. Find the probability that a sample of 26 randomly selected adult males has a mean weight greater than 150 lb. c. What do you conclude about the safety of this​ elevator?

a. greater than 150lb is 0.8729 b. 26 males with mean greater than 150lb is 0.9999 c. does this elevator appear safe? No, because there is a good chance that 26 randomly selected adult male passengers will exceed the elevator capacity.

The ages​ (years) of three government officials when they died in office were 55​, 44​, and 61. Complete parts​ (a) through​ (d).

explaination - finding ranges, find the mean of all the sample ranges and compare to the calculated x-bar c. Compare the population range to the mean of the sample ranges. Choose the correct answer below. -The population range is not equal to the mean of the sample ranges​ (it is also not equal to the age of the oldest official or age of the youngest official at the time of​ death). d. Do the sample ranges target the value of the population​ range? In​ general, do sample ranges make good estimators of population​ ranges? Why or why​ not? - The sample ranges do not target the population​ range, therefore, sample ranges do not make good estimators of population ranges.

Finding probabilities associated with distributions that are standard normal distributions is equivalent to​ _______.

finding the area of the shaded region representing that probability.

Density Curve

graph of any continuous probability distribution

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is between −2.12 and 3.81 and draw a sketch of the region.

sketch is the fill in between the numbers math is area of 3.81 (0.9999) minus the area of -2.12 (0.0170) make sure not to fat finger = 0.9829

finding an area range with Z scores

subtract highest Z-score area minus the Lower Z-score area (the large number minus small)

states that if the sample size is large​ enough, the distribution of sample means can be approximated by a normal​ distribution, even if the original population is not normally distributed.

the central limit theorem

Which of the following is NOT a descriptor of a normal distribution of a random​ variable?

the graph is centered around 0

Which of the following does NOT describe the standard normal​ distribution?

the graph is uniform

uniform distribution

values equally spread over the range of possible values - Rectangular graphs

Find X from Z score

x = mean + (Z X standard deviation) when finding the area above the X make sure to subtract the area from 1 to find the correct Z score, if needing to find that area, make sure to take 1 and subtract the table value for Z score table always shows the area to the left, or less than value

if the mean length of newborn infants is 52.5cm, and the standard deviation is 4.5cm, and a sample of 15 newborns are randomly selected. what is the probability that the mean length is greater than 56cm?

z = [x (bar) - μ(of xbar)] / σ(of xbar) = (x(bar) - μ) / (σ / √n) = [(xbar - μ) / 1] X (√n / σ) = [(xbar - μ) / σ ] n = 15 σ = 4.5cm μ = 52.5cm p(xbar > 56) = P(z>3.0123) - - - z= (56-52.5)/4.5= 0.777 now multiply by √15 = 3.0123 = Z the probability of P(z) = 0.9987 so the probability of P(z>3.0123) = 1- 0.9987 SO P(z>3.0123) = 0.0013 VERY SMALL

Z score

z = x - mean / standard deviation

The standard deviation of the distribution of sample means is

σ / √n

Which of the following is NOT a requirement for a density​ curve?

​"The graph is centered around​ 0"

Which of the following groups of terms can be used interchangeably when working with normal​ distributions?

​areas, probability, and relative frequencies