You are watching: What requirements are necessary for a normal probability distribution to be a standard

updated 6 years earlier by GreenHero64

grade levels: College: very first year, College: second year, College: 3rd year, College: fourth year

Subjects:statistics, mathematics, probability & statistics

show moreless

Page to share: collection (this page) Cards Quiz matching Bingo print

COPY

Embed this setcancel

COPY

password changes based on your size an option

Size: little (450 x 345) tool (600 x 460) huge (750 x 575) Custom dimension

X Maintain element ratio

Show:

all Cards

**57**

All57 Un-marked significant

note all un-mark all

1

What needs are essential for a regular probability distribution to it is in a *standard* typical probability distribution?

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

2

The wait times between a subway leave schedule and also the arrival of a passenger space uniformly distributed between 0 and 8 minutes. Discover the probability that a randomly selected passenger has a waiting time less than 0.75 minutes.

* * **0.098**

*P(less than 0.75) = (length that shaded region) x (height the shaded region)*

*= (0.75 – 0) x (1 ÷ 8) *round come 2 dec. Places*

*=(0.75) x (0.13) = 0.0975*

3

Find the area that the shaded region. The graph depicts the conventional normal distribution with median 0 and also standard deviation 1.

The area the the shaded an ar is **0.7019**.

4

Find the area the the shaded region. The graph depicts the traditional normal distribution of bone thickness scores with average 0 and also standard deviation 1.

The area of the shaded an ar is ** 0.7309**.

*Area = (0.8944 – 0.1635)*

5

Find the suggested z score. The graph depicts the conventional normal distribution with average 0 and also standard deviation 1.

The suggested z score is ** -0.74**.

6

Find the indicated z score. The graph depicts the traditional normal distribution with mean 0 and also standard deviation 1.

The indicated z score is **1.05**.

*Symmetric to Area the 0.1469 (z score that -1.05)*

*OR*

*Area = 1 – 0.1469*

7

Assume the a randomly selected topic is offered a bone thickness test. Those check scores are normally dispersed with a mean of 0 and a typical deviation that 1.

Find the probability that a offered score is less than 1.66 and draw a lay out of the region.

The probability is ** 0.9515**.

8

Assume the readings on thermometers are normally spread with a mean of 0°C and also a standard deviation of 1.00°C.

Find the probability the a randomly selected thermometer reads greater than –1.78 and also draw a map out of the region.

The probability is ** 0.9625**.

*Symmetric to z-score that 1.78 (Area = 0.9625)*

*OR*

*1 – 0.0375*

9

Assume the readings ~ above thermometers room normally spread with a mean of 0°C and a standard deviation the 1.00°C.

Find the probability the a randomly selected thermometer reads between –1.03 and –0.02 and draw a lay out of the region.

The probability is ** 0.3405**.

*Area = (0.4920 – 0.1515)*

10

Assume that a randomly selected subject is provided a bone density test. Those test scores space normally dispersed with a average of 0 and also a standard deviation the 1.

Find the probability that a offered score is in between –2.02 and also 3.84 and draw a sketch of the region.

The probability is **0.9782**.

** For worths of z over 3.48, usage 0.9999*

*Area = (0.9999 – 0.0217)*

11

Assume the the readings top top the thermometers are normally distributed with a typical of 0° and standard deviation of 1.00°C. A thermometer is randomly selected and tested.

Draw a sketch and find the temperature reading matching to P97, the 97th percentile. This is the temperature analysis separating the bottom 97% from the height 3%.

The temperature because that P97 is around ** 1.88 °**.

*Area the 0.9700 ≈ 0.9699 (z score the 1.88)*

12

Assume the the readings top top the thermometers room normally dispersed with a typical of 0° and also standard deviation of 1.00°C. I think 2.1% of the thermometers room rejected since they have actually readings that are too high and also another 2.1% are rejected because they have actually readings that are too low.

Draw a sketch and also find the two readings that are cutoff values separating the rubbish thermometers indigenous the others.

The cutoff values space ** –2.03, 2.03** degrees.

*Area values nearest 0.021 and 0.979*

*≈ 0.0212 and 0.9788*

13

Find the shown area under the curve that the traditional normal distribution, then transform it come a percentage and fill in the blank.

About **72.86**% the the area is between z = –1.1 and also z = 1.1 (or within 1.1 conventional deviations the the mean).

14

Which of the adhering to groups the terms have the right to be provided interchangeably as soon as working v normal distributions?

**areas, probability, and relative frequencies**

*Area, probability, percentages, and relative frequencies space all offered interchangeably when working through normal distributions.*

15

Which the the complying with is no a necessity for a density curve?

** The graph is centered about 0. **

*"The graph is focused around 0" is not a requirement for a density curve.*

16

Which that the complying with does NOT describe the standard normal distribution?

**The graph is uniform.**

*The typical normal distribution is no uniform due to the fact that its graph is bell-shaped.*

17

Which that the complying with would be info in a question asking friend to uncover the area that a an ar under the conventional normal curve as a solution?

** A distance on the horizontal axis is provided **

*When offered a z-score, you are usually detect the area the the shaded region under the typical normal curve. For the standard normal curve, a z-score is a distance along the horizontal axis.*

18

Assume the a randomly selected topic is given a bone thickness test. Those check scores space normally dispersed with a average of 0 and also a traditional deviation the 1.

Find the probability the a given score is much less than 3.61 and draw a map out of the region.

The probability is **0.9999**.

19

Assume the a randomly selected subject is provided a bone density test. Bone density test scores room normally spread with a average of 0 and a traditional deviation the 1.

Draw a graph and also find P3, the third percentile. This is the bone density score separating the bottom 3% from the optimal 97%.

The bone thickness score equivalent to upper P 3 is **–1.88**.

*Area = 0.0300 ≈ 0.0301*

20

Find the indicated an important value.

*z0.04*

z0.04 = **1.75**

*Area = 0.0400 ≈ 0.0401*

21

A survey uncovered that women"s heights are normally spread with median 63.6 in and also standard deviation 2.3 in. A branch of the army requires women"s heights come be between 58 in and 80 in.

find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch that the military because they space too quick or too tall? If this branch of the military alters the elevation requirements so the all women space eligible other than the shortest 1% and also the tallest 2%, what room the new height requirements? **a.** The portion of ladies who meet the elevation requirement is ** 99.25**%.

**(58 – 63.6) ÷ 2.3 = -2.43 (min)*

**(80 – 63.6) ÷ 2.3 = 7.13 (max)*

P(-2.43 **58.2** in and at many **68.3** in.

**shortest 1% = 0.0100 ≈ 0.0099 (value that -2.33)*

**tallest 2% = 0.9800 ≈ 0.9798 (value the 2.05)*

** *µ + (x × σ)**

*63.6 + (-2.33 × 2.3) = 58.241*

*63.6 + (2.05 × 2.3) = 68.315*

22

Men"s heights room normally distributed with average 68.7 in and standard deviation that 2.8 in. Women"s heights are normally spread with mean 63.5 in and standard deviation of 2.5 in. The standard doorway elevation is 80 in.

What portion of males are too tall to fit v a conventional doorway without bending, and also what portion of women space too tall to fit through a typical doorway without bending? If a statistician designs a home so that every one of the doorways have heights that are sufficient for all men other than the tallest 5%, what doorway height would be used? **a.** The portion of men who are too tall to fit v a standard door without bending is ** 0**%.

**(80 – 68.7) ÷ 2.8 = 4.035714286 ≈ 4.04*

*P(z > 4.04)*

*= 1 – P(z 0%.*

**(80 – 63.5) ÷ 2.5 = 6.6*

*P(z > 6.60)*

*= 1 – P(z 73.3 in.*

23

**tallest 5% = 0.9500 ≈ 1.645*

** µ + (x × σ)**

*68.7 + (1.645 × 2.8) = 73.306*

23

The lengths the pregnancies are normally spread with a average of 266 days and also a traditional deviation of 15 days.

discover the probability that a pregnant lasting 307 days or longer. If the size of pregnancy is in the shortest 2%, then the baby is premature. Uncover the length that separates premature babies indigenous those who room not premature. **a.** The probability the a pregnancy will certainly last 307 work or longer is ** 0.0032**.

**(307 – 266) ÷ 15 = 2.7333333 (z score)*

*P(z > 2.73)*

*= 1 – P(z 235 job are taken into consideration premature.*

24

**lowest 2% = 0.0200 ≈ 0.0202 = -2.05*

** µ + (x × σ)**

*266 + (-2.05 × 15) = 235.25*

24

Assume the the Richter range magnitudes the earthquakes are normally dispersed with a typical of 1.087 and also a conventional deviation the 0.562.

Earthquakes through magnitudes much less than 2.000 space considered "microearthquakes" that are not felt. What percentage of earthquakes autumn into this category? Earthquakes above 4.0 will reason indoor items come shake. What percent of earthquakes loss into this category? discover the 95th percentile. Will all earthquakes above the 95th percentile reason indoor item to shake? **a.** ** 94.74**%

**(2.00 – 1.087) ÷ 0.562 = 1.62455516*

*P(z 0%*

**(4.00 – 1.087) ÷ 0.562 = 5.183274021*

*P(z > 5.18)*

*= 1 – P(z 2.011*

25

26

27

**95th = 0.9500 = 1.645 (z score)*

** µ + (x × σ)**

*1.087 + (1.645 × 0.562) = 2.01149*

**d.** No, due to the fact that not all earthquakes over the 95th percentile have magnitudes over 4.0.

25

Chocolate chip cookies have actually a circulation that is roughly normal v a average of 23.1 coco chips per cookie and also a standard deviation the 2.6 chocolate chips every cookie.

Find P5 and also P95.

How might those values be helpful to the producer of the chocolate chip cookies?

P5 = **18.8**

**P5 = 0.0500 = -1.645 (z score)*

** µ + (x × σ)**

*23.1 + (-1.645 × 2.6) = 18.823*

P95 = **27.4**

**95th = 0.9500 = 1.645 (z score)*

** µ + (x × σ)**

*23.1 + (1.645 × 2.6) = 27.377*

**The values have the right to be supplied to identify cookies with an unusually low or high number of chocolate chips, so those numbers can be used to monitor the production procedure to ensure that the numbers of coco chips stays within reasonable limits.**

26

Pulse prices of women are normally dispersed with a typical of 77.5 beats per minute and also a conventional deviation of 11.6 beats every minute.

What are the worths of the mean and also standard deviation after converting all pulse rates of women to z scores using z = (x – **µ**) ÷ **σ**?

The original pulse prices are measure with units of "beats per minute". What are the devices of the equivalent z scores?

**µ = 0**

**σ = 1**

**The z scores room numbers without systems of measurement.**

27

Find the area that the shaded region. The graph come the appropriate depicts IQ scores of adults, and those scores space normally distributed with a typical of 100 and also a traditional deviation that 15.

The area of the shaded region is ** 0.3707**.

**z = (95 – 100) ÷ 15 = -0.333*

*P(z *

*The probability that a randomly selected adult has an IQ less than 118 is 0.8849.*

**z = (118 – 100) ÷ 15 = 1.2*

*P(z 3, i m sorry is the IQ score separating the bottom 3% indigenous the height 97%.*

Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and also those scores room normally distributed with a median of 100 and also a conventional deviation the 15.

Which of the complying with is not true?

A z-score is one area under the typical curve. A z-score is a conversion the standardizes any value indigenous a normal distribution to a standard normal distribution. The area in any type of normal distribution bounded by part score x is the same as the area bounded through the equivalent z-score in the traditional normal distribution. If values room converted to standard z-scores, then steps for working through all normal distributions space the very same as those for the traditional normal distribution.What problems would produce a negative z-score?

Which the the complying with statistics room unbiased estimators that population parameters?

Sample proportion provided to calculation a population proportion. Sample range used to estimate a population range. Sample variance used to estimate a population variance. Sample typical used to estimate a population mean. Sample conventional deviation supplied to calculation a populace standard deviation. Sample typical used to calculation a population median.The assets (in billions of dollars) the the four wealthiest human being in a certain country space 37, 27, 22, 19. Assume that samples of dimension n = 2 space randomly selected through replacement native this population of four values.

After identifying the 16 different possible samples and also finding the mean of each sample, construct a table representing the sampling distribution of the sample mean. In the table, values of the sample mean that are the same have been combined. Compare the median of the population to the average of the sampling distribution of the sample mean. Do the sample means target the value of the population mean? In general, perform sample means make good estimates that population means? Why or why not? ** a. (see picture)**

*37-37; 37-27; 37-22; 37-19; 27-37; 27-27; 27-22; 27-19; 22-37; 22-27; 22-22; 22-19; 19-37; 19-27; 19-22; 19-19 – (16 samples)*

*The distinct way are 37, 32, 29.5, 28, 27, 24.5, 23, 22, 20.5, and also 19. To uncover the corresponding probability for each mean, division the number of occurrences the that median by the total number of samples.*

**b. **The average of the population, **26.25**, is **equal to** the average of the sample means, ** 26.25**.

*Recall that the mean of a collection of data is the measure up of center found by adding the data values and also dividing the total by the number of data values.*

*(37, 27, 22, 19) ÷ 4 = 26.25*

**c.** The sample method **target** the populace mean. In general, sample means **do** make good estimates that population means because the mean is ** one unbiased** estimator.

Which of the following is not a property of the sampling circulation of the sample mean?

The sample means target the value of the population mean. The circulation of the sample mean tends come be skewed to the appropriate or left. The median of the sample way is the populace mean. The supposed value the the sample typical is equal to the populace mean.Which the the following is a biased estimator? That is, i m sorry of the complying with does no target the population parameter?

The populace of current statistics students has ages with median **µ **and typical deviation **σ**. Samples of statistics students space randomly selected so that there are precisely 37 student in every sample. For each sample, the mean period is computed.

What does the main limit theorem call us around the distribution of those mean ages?

Which the the adhering to is no a conclusion that the main Limit Theorem?

The traditional deviation of every sample means is the population standard deviation split by the square root of the sample size. The typical of all sample method is the population mean**µ**. The distribution of the sample method**x̄**will, together the sample size increases, strategy a regular distribution. The distribution of the sample data will method a normal circulation as the sample size increases.Which that the complying with is not a typically used practice?

If the circulation of the sample way is normally distributed, and also n > 30, then the populace distribution is typically distributed. The distribution of sample method gets closer to a normal distribution as the sample dimension n it s okay larger. If the original population is not typically distributed and also n > 30, the circulation of the sample way can it is in approximated sensibly well by a common distribution. If the original populace is normally distributed, then for any sample size n, the sample way will be typically distributed.Assume that women"s heights are normally distributed with a mean given by **µ** = 63.6 in, and also a traditional deviation provided by** σ** = 2.6 in.

**(a)** If 1 mrs is randomly selected, discover the probability that her height is less than 64 in.

**(b)** If 44 women are randomly selected, find the probability that they have actually a mean elevation less 보다 64 in.

Assume that women"s heights are normally distributed with a mean given by **µ** = 62.2 in, and also a typical deviation offered by **σ** = 2.8 in.

Women have actually head circumferences that are normally distributed with a mean given by **µ** = 24.27 in., and a traditional deviation given by** σ** = 0.8 in.

A ski gondola dead skiers to the optimal of a mountain. It bear a plaque stating that the maximum volume is 12 people or 1920 lb. The capacity will be gone beyond if 12 world have weights through a mean higher than (1920 lb)/12 = 160 lb. Assume the weights of passengers space normally dispersed with a median of 177.5 lb and also a standard deviation of 40.4 lb.

find the probability the if an individual passenger is randomly selected, their weight will certainly be greater than 160 lb. Uncover the probability that 12 randomly selected passengers will have actually a typical weight the is greater than 160 lb (so the their complete weight is greater than the gondola maximum capacity of 1920 lb)An technician is going come redesign an ejection seat because that an airplane. The seat was designed for pilots weighing between 150 lb and also 191 lb. The brand-new population the pilots has normally dispersed weights with a median of 156 lb and also a traditional deviation of 25.7 lb.

If a pilot is randomly selected, find the probability that his weight is between 150 lb and 191 lb. If 33 various pilots space randomly selected, uncover the probability that their average weight is between 150 lb and 191 lb. As soon as redesigning the ejection seat, i beg your pardon probability is more relevant?An airliner tote 200 passengers and also has doors with a height of 72 in. Heights of males are normally spread with a mean of 69.0 in and a conventional deviation the 2.8 in.

If a masculine passenger is randomly selected, find the probability the he deserve to fit v the doorway there is no bending. If fifty percent of the 200 passengers are men, find the probability that the mean height of the 100 men is much less than 72 in. When considering the comfort and also safety of passengers, which result is more relevant: the probability indigenous part (a) or the probability from part (b)? Why? once considering the comfort and also safety of passengers, why are women ignored in this case?Before every flight, the pilot should verify that the complete weight the the load is much less than the maximum allowable fill for the aircraft. The aircraft can bring 35 passengers, and also a flight has fuel and baggage that allows for a full passenger pack of 5,880 lb. The pilot sees the the plane is full and also all passengers are men. The aircraft will be overloaded if the average weight the the passengers is greater than (5,880 lb)/35 = 168 lb.

What is the probability the the plane is overloaded?

Should the pilot take it any activity to correct for an overloaded aircraft? Assume that weights of males are normally distributed with a average of 175.5 lb and a typical deviation of 38.7.

The worth given listed below is discrete. Use the continually correction and also describe the an ar of the normal distribution that coincides to the shown probability.

See more: How Old Was Steve Martin When Did Steve Martin Hair Go Grey, ! Steve Martin Didn'T Always Have Grey Hair

Probability of specifically 5 passenger who carry out not present up because that a flight

If np ≥ 5 and nq ≥ 5, estimate P(more than 5) v n = 12 and p = 0.4 by utilizing the normal distribution as one approximation come the binomial distribution; if np

**The normal distribution cannot it is in used.**

*np = (12)(0.4) = 4.8 *

If np ≥ 5 and also nq ≥ 5, estimate P(more 보다 5) through n = 13 and also p = 0.6 by using the normal distribution as an approximation to the binomial distribution; if np

P(at least 8) = **0.567**