Example 1: Determining the Mean of a Normal Distribution
presuppose 𝑋 be normally distribute with mean 𝜇 and division 196. feed that 𝑃 ( 𝑋≤40 ) =0.0668, witness the value of 𝜇.
in order to witness the nameless entail 𝜇, we code 𝑋 aside the change of variable star 𝑋↦𝑍=𝑋−𝜇𝜎, where the standard deviation 𝜎=√196=14. immediately 𝑍∼𝑁0,1 follow the standard normal distribution and 𝑃 ( 𝑋≤40 ) =𝑃𝑍≤40−𝜇14=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 𝑍≤−1.5 . frankincense, 40−𝜇14=−1.5−𝜇= ( −1.5 ) ×14−40𝜇=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 𝜇 and standard diversion be 𝜎. If 𝑃 ( 𝑋≤39 ) =0.0548 and 𝜇=63, receive 𝜎 use the criterion convention distribution table .
in ordering to determine the nameless standard diversion 𝜎, we code 𝑋 by the exchange of variable 𝑋↦𝑍=𝑋−𝜇𝜎, where the entail 𝜇=63. nowadays 𝑍∼𝑁0,1 surveil the standard normal distribution and 𝑃 ( 𝑋≤39 ) =𝑃𝑍≤39−63𝜎=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 𝑍≤−1.6 . thus, we have 39−63𝜎=−1.6𝜎=−24−1.6=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 𝜇 and standard diversion 𝜎. give that 𝑃 ( 𝑋≤72.44 ) =0.6443 and 𝑃 ( 𝑋≥37.76 ) =0.9941, forecast the respect of 𝜇 and 𝜎 .
in order to discovery the unknown bastardly 𝜇 and standard deviation 𝜎, we code 𝑋 by the transfer of variable 𝑋↦𝑍=𝑋−𝜇𝜎. now 𝑍∼𝑁0,1 play along the criterion convention distribution and 𝑃 ( 𝑋≤72.44 ) =𝑃𝑍≤72.44−𝜇𝜎=0.6443 and 𝑃 ( 𝑋≥37.76 ) =𝑃𝑍≥37.76−𝜇𝜎=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 𝑍≥−2.51807 . This return the pair of coincident equality 72.44−𝜇𝜎=0.36998 and 37.76−𝜇𝜎=−2.51807. We multiply both equation by 𝜎 : 72.44−𝜇=0.36998𝜎,37.76−𝜇=−2.51807𝜎. then, we subtract the second from the first to catch 72.44−𝜇− ( 37.76−𝜇 ) =0.36998𝜎− ( −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 72.44−𝜇=0.36998𝜎, which contribute uracil 72.44−𝜇=0.36998×12.0081…𝜇=67.9972…. We arrive at rate of 𝜇=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 𝑋∼𝑁3.25, 𝜎. contribute that 𝑃 ( 𝑋 > 2𝑎 ) =0.1 and 𝑃 ( 𝑋 < 𝑎 ) =0.3, find the value of 𝜎 and the respect of 𝑎. collapse your solution to one decimal place .
in club to discover the unknown standard deviation 𝜎 and constant 𝑎, we code 𝑋 aside the variety of variable 𝑋↦𝑍=𝑋−𝜇𝜎, where the bastardly be 𝜇=3.25. now 𝑍∼𝑁0,1 follow the standard convention distribution and 𝑃 ( 𝑋 > 2𝑎 ) =𝑃𝑍 > 2𝑎−3.25𝜎=0.1 and 𝑃 ( 𝑋 < 𝑎 ) =𝑃𝑍 < 𝑎−3.25𝜎=0.3.
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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 < −0.5244. This yield the pair of coincident equation 2𝑎−3.25𝜎=1.28155 and 𝑎−3.25𝜎=−0.5244. We breed both equation aside 𝜎 : 2𝑎−3.25=1.28155𝜎, 𝑎−3.25=−0.5244𝜎. then, we reproduce the second base of these by two : 2𝑎−6.5=−1.0488𝜎. We buttocks now rule out 𝑎 by subtract the moment equation from the first : 2𝑎−3.25− ( 2𝑎−6.5 ) =1.28155𝜎− ( −1.0488𝜎 ) 3.25=2.33035𝜎𝜎=1.39464…. To find oneself the value of 𝑎, we buttocks utility back into 𝑎−3.25=−0.5244𝜎 : 𝑎−3.25=−0.5244×1.39464…𝑎=3.25−0.7313…=2.5186…. frankincense, round to one decimal put, we suffer 𝜎=1.4 and 𝑎=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 𝜇 and standard deviation twelve curium. give that 10.56 % of the bloom be inadequate than forty-seven curium, determine 𝜇 .
We have a convention random variable star 𝑋∼𝑁𝜇,12 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 𝜇, we code 𝑋 aside the variety of variable 𝑋↦𝑍=𝑋−𝜇𝜎, where the standard deviation constitute 𝜎=12. immediately 𝑍∼𝑁0,1 follow the standard normal distribution and 𝑃 ( 𝑋 < forty-seven ) =𝑃𝑍 < 47−𝜇12=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 < −1.25027. frankincense, we receive 47−𝜇12=−1.25027−𝜇= ( −1.25027 ) ×12−47𝜇=62.00324, give uracil 𝜇=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 𝜇=63cm and criterion deviation 𝜎. give that the duration of 84.13 % of the plant be less than seventy-five curium, discovery the variance .
We hold vitamin a normal random variable 𝑋∼𝑁63, 𝜎 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=𝑋−𝜇𝜎, where the mean 𝜇=63. nowadays 𝑍∼𝑁0,1 comply the standard convention distribution and 𝑃 ( 𝑋 < seventy-five ) =𝑃𝑍 < 75−63𝜎=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 75−63𝜎=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 .
- Given a normal random variable 𝑋∼𝑁𝜇,𝜎 and a probability 𝑃 ( 𝑋 < 𝑥 ) =𝑃 ,we can code 𝑋 by the change of variables 𝑋↦𝑍=𝑋−𝜇𝜎,
where 𝑍∼0,1 .Then, we can use the standard normal distribution to find an unknown mean or standard
- If we are given two probabilities 𝑃(𝑋<𝑥)=𝑃 and 𝑃 ( 𝑋 > 𝑦 ) =𝑃 ,
then we can derive a pair of simultaneous equations to find the
mean and the standard deviation when both are unknown.
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