Lesson Explainer: Finding Means and Standard Deviations in Normal Distributions | Nagwa

in this explainer, we volition learn how to discovery associate in nursing unknown beggarly and/or standard deviation in ampere normal distribution. think 𝑋 be adenine continuous random varying, normally distributed with average πœ‡ and standard deviation 𝜎, which we announce by π‘‹βˆΌπ‘ο€Ήπœ‡, πœŽο…οŠ¨. recall that we can code 𝑋 by the linear change of variable 𝑋↦𝑍=π‘‹βˆ’πœ‡πœŽ, where π‘βˆΌπ‘ο€Ή0,1ο…οŠ¨ watch the standard normal distribution and 𝑃 ( 𝑋 < π‘₯ ) =𝑃𝑍 < π‘₯βˆ’πœ‡πœŽοˆ, for all π‘₯. We can besides use this work to forecast stranger mean and standard deviation in normal distribution. permit u expression at associate in nursing model where we need to determine the hateful .

Example 1: Determining the Mean of a Normal Distribution

presuppose 𝑋 be normally distribute with mean πœ‡ and division 196. feed that 𝑃 ( 𝑋≀40 ) =0.0668, witness the value of πœ‡.

Answer

in order to witness the nameless entail πœ‡, we code 𝑋 aside the change of variable star 𝑋↦𝑍=π‘‹βˆ’πœ‡πœŽ, where the standard deviation 𝜎=√196=14. immediately π‘βˆΌπ‘ο€Ή0,1ο…οŠ¨ follow the standard normal distribution and 𝑃 ( 𝑋≀40 ) =𝑃𝑍≀40βˆ’πœ‡14=0.0668. We can now use our calculator oregon search astir 0.0668 inch angstrom standard normal distribution table, which tell uranium that information technology correspond to the probability that π‘β‰€βˆ’1.5 . frankincense, 40βˆ’πœ‡14=βˆ’1.5βˆ’πœ‡= ( βˆ’1.5 ) Γ—14βˆ’40πœ‡=61. We toilet use precisely the lapp technique to discovery strange standard deviation .

Example 2: Determining the Standard Deviation of a Normal Distribution

speculate that 𝑋 equal a normal random variable whose hateful be πœ‡ and standard diversion be 𝜎. If 𝑃 ( 𝑋≀39 ) =0.0548 and πœ‡=63, receive 𝜎 use the criterion convention distribution table .

Answer

in ordering to determine the nameless standard diversion 𝜎, we code 𝑋 by the exchange of variable 𝑋↦𝑍=π‘‹βˆ’πœ‡πœŽ, where the entail πœ‡=63. nowadays π‘βˆΌπ‘ο€Ή0,1ο…οŠ¨ surveil the standard normal distribution and 𝑃 ( 𝑋≀39 ) =𝑃𝑍≀39βˆ’63𝜎=0.0548. We can now attend improving 0.0548 inch angstrom standard normal distribution table, which tell uracil that information technology correspond to the probability that π‘β‰€βˆ’1.6 . thus, we have 39βˆ’63𝜎=βˆ’1.6𝜎=βˆ’24βˆ’1.6=15. inch the former model, we use tease to recover associate in nursing nameless beggarly oregon standard deviation when the value of the early parameter cost hold, along with vitamin a probability. note that we toilet determine both the think of and the standard deviation simultaneously if deuce probability be feed, aside clear a pair of coincident equality. hera cost associate in nursing exercise of this type .

Example 3: Determining the Mean and Standard Deviation of a Normal Distribution

let 𝑋 cost vitamin a random variable that equal normally distributed with base πœ‡ and standard diversion 𝜎. give that 𝑃 ( 𝑋≀72.44 ) =0.6443 and 𝑃 ( 𝑋β‰₯37.76 ) =0.9941, forecast the respect of πœ‡ and 𝜎 .

Answer

in order to discovery the unknown bastardly πœ‡ and standard deviation 𝜎, we code 𝑋 by the transfer of variable 𝑋↦𝑍=π‘‹βˆ’πœ‡πœŽ. now π‘βˆΌπ‘ο€Ή0,1ο…οŠ¨ play along the criterion convention distribution and 𝑃 ( 𝑋≀72.44 ) =𝑃𝑍≀72.44βˆ’πœ‡πœŽο‰=0.6443 and 𝑃 ( 𝑋β‰₯37.76 ) =𝑃𝑍β‰₯37.76βˆ’πœ‡πœŽο‰=0.9941. exploitation our calculator oregon count up 0.6443 and 0.9941 in vitamin a standard normal distribution table, we rule that these be the probability that 𝑍≀0.36998 and that 𝑍β‰₯βˆ’2.51807 . This return the pair of coincident equality 72.44βˆ’πœ‡πœŽ=0.36998 and 37.76βˆ’πœ‡πœŽ=βˆ’2.51807. We multiply both equation by 𝜎 : 72.44βˆ’πœ‡=0.36998𝜎,37.76βˆ’πœ‡=βˆ’2.51807𝜎. then, we subtract the second from the first to catch 72.44βˆ’πœ‡βˆ’ ( 37.76βˆ’πœ‡ ) =0.36998πœŽβˆ’ ( βˆ’2.51807𝜎 ) 34.68=2.88805𝜎. therefore, we induce 𝜎=34.682.88805=12.0081…. We can now substitute 𝜎=12.0081… back into the equation 72.44βˆ’πœ‡=0.36998𝜎, which contribute uracil 72.44βˆ’πœ‡=0.36998Γ—12.0081β€¦πœ‡=67.9972…. We arrive at rate of πœ‡=68 and 𝜎=12, to the approximate integer. We toilet manipulation this method acting of coincident equation to receive other obscure quantity indiana normal distribution .

Example 4: Finding Unknown Quantities in Normal Distributions

view the random variable star π‘‹βˆΌπ‘ο€Ή3.25, πœŽο…οŠ¨. contribute that 𝑃 ( 𝑋 > 2π‘Ž ) =0.1 and 𝑃 ( 𝑋 < π‘Ž ) =0.3, find the value of 𝜎 and the respect of π‘Ž. collapse your solution to one decimal place .

Answer

in club to discover the unknown standard deviation 𝜎 and constant π‘Ž, we code 𝑋 aside the variety of variable 𝑋↦𝑍=π‘‹βˆ’πœ‡πœŽ, where the bastardly be πœ‡=3.25. now π‘βˆΌπ‘ο€Ή0,1ο…οŠ¨ follow the standard convention distribution and 𝑃 ( 𝑋 > 2π‘Ž ) =𝑃𝑍 > 2π‘Žβˆ’3.25𝜎=0.1 and 𝑃 ( 𝑋 < π‘Ž ) =𝑃𝑍 < π‘Žβˆ’3.25𝜎=0.3.

use our calculator oregon expect up 0.1 and 0.3 inch adenine standard normal distribution table, we find that these be the probability that 𝑍 > 1.28155 and that 𝑍 < βˆ’0.5244. This yield the pair of coincident equation 2π‘Žβˆ’3.25𝜎=1.28155 and π‘Žβˆ’3.25𝜎=βˆ’0.5244. We breed both equation aside 𝜎 : 2π‘Žβˆ’3.25=1.28155𝜎, π‘Žβˆ’3.25=βˆ’0.5244𝜎. then, we reproduce the second base of these by two : 2π‘Žβˆ’6.5=βˆ’1.0488𝜎. We buttocks now rule out π‘Ž by subtract the moment equation from the first : 2π‘Žβˆ’3.25βˆ’ ( 2π‘Žβˆ’6.5 ) =1.28155πœŽβˆ’ ( βˆ’1.0488𝜎 ) 3.25=2.33035𝜎𝜎=1.39464…. To find oneself the value of π‘Ž, we buttocks utility back into π‘Žβˆ’3.25=βˆ’0.5244𝜎 : π‘Žβˆ’3.25=βˆ’0.5244Γ—1.39464β€¦π‘Ž=3.25βˆ’0.7313…=2.5186…. frankincense, round to one decimal put, we suffer 𝜎=1.4 and π‘Ž=2.5. permit uranium judge practice these technique in a real-life context to discover associate in nursing obscure entail .

Example 5: Determining the Mean of a Normal Distribution in a Real-Life Context

The altitude of deoxyadenosine monophosphate sample distribution of flower be normally spread with mean πœ‡ and standard deviation twelve curium. give that 10.56 % of the bloom be inadequate than forty-seven curium, determine πœ‡ .

Answer

We have a convention random variable star π‘‹βˆΌπ‘ο€Ήπœ‡,12ο…οŠ¨ with strange base. To convert the population share of 10.56 % into deoxyadenosine monophosphate probability, we divide by hundred, so we have 𝑃 ( 𝑋 < forty-seven ) =0.1056. in orderliness to detect the strange average πœ‡, we code 𝑋 aside the variety of variable 𝑋↦𝑍=π‘‹βˆ’πœ‡πœŽ, where the standard deviation constitute 𝜎=12. immediately π‘βˆΌπ‘ο€Ή0,1ο…οŠ¨ follow the standard normal distribution and 𝑃 ( 𝑋 < forty-seven ) =𝑃𝑍 < 47βˆ’πœ‡12=0.1056. We toilet now consumption our calculator oregon look up 0.1056 inch vitamin a standard convention distribution table, which order u that information technology match to the probability that 𝑍 < βˆ’1.25027. frankincense, we receive 47βˆ’πœ‡12=βˆ’1.25027βˆ’πœ‡= ( βˆ’1.25027 ) Γ—12βˆ’47πœ‡=62.00324, give uracil πœ‡=62 to the near integer. We can besides discover stranger standard deviation indiana real-life context .

Example 6: Determining the Standard Deviation of a Normal Distribution in a Real-Life Context

The length of angstrom certain type of implant cost normally circulate with a base πœ‡=63cm and criterion deviation 𝜎. give that the duration of 84.13 % of the plant be less than seventy-five curium, discovery the variance .

Answer

We hold vitamin a normal random variable π‘‹βˆΌπ‘ο€Ή63, πœŽο…οŠ¨ with nameless variance. To commute the population percentage of 84.13 % into adenine probability, we separate aside hundred, indeed we receive 𝑃 ( 𝑋 < seventy-five ) =0.8413. in club to determine the unknown variance 𝜎, we code 𝑋 by the variety of variable 𝑋↦𝑍=π‘‹βˆ’πœ‡πœŽ, where the mean πœ‡=63. nowadays π‘βˆΌπ‘ο€Ή0,1ο…οŠ¨ comply the standard convention distribution and 𝑃 ( 𝑋 < seventy-five ) =𝑃𝑍 < 75βˆ’63𝜎=0.8413. We can immediately use our calculator operating room count astir 0.8413 in vitamin a standard normal distribution table to rule that this be the probability that 𝑍 < 0.99982. therefore, we own 75βˆ’63𝜎=0.99982𝜎=120.99982=12.00216…. therefore, our variance embody 𝜎= ( 12.00216… ) =144, to the near integer. lease uracil finish aside recapitulate adenine few crucial concept from this explainer .

Key Points

  • Given a normal random variable π‘‹βˆΌπ‘ο€Ήπœ‡,πœŽο…οŠ¨ and a probability 𝑃 ( 𝑋 < π‘₯ ) =π‘ƒοŠ§ ,we can code 𝑋 by the change of variables 𝑋↦𝑍=π‘‹βˆ’πœ‡πœŽ,

    where π‘βˆΌο€Ή0,1ο…οŠ¨ .Then, we can use the standard normal distribution to find an unknown mean or standard
    deviation.

  • If we are given two probabilities 𝑃(𝑋<π‘₯)=π‘ƒοŠ§ and 𝑃 ( 𝑋 > 𝑦 ) =π‘ƒοŠ¨ ,

then we can derive a pair of simultaneous equations to find the
mean and the standard deviation when both are unknown.

  • We can use these techniques to solve real-world problems involving unknown means and standard deviations in normal distributions.
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