**code**π by the linear change of variable πβ¦π=πβππ, where πβΌποΉ0,1ο ο¨ watch the standard normal distribution and π ( π < π₯ ) =ποΌπ < π₯βππο, 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 π be normally distribute with mean π and division 196. feed that π ( πβ€40 ) =0.0668, witness the value of π.

## Answer

in order to witness the nameless entail π, we code π aside the change of variable star πβ¦π=πβππ, where the standard deviation π=β196=14. immediately πβΌποΉ0,1ο ο¨ follow the standard normal distribution and π ( πβ€40 ) =πο½πβ€40βπ14ο=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 πβ€β1.5 . frankincense, 40βπ14=β1.5βπ= ( β1.5 ) Γ14β40π=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 π equal a normal random variable whose hateful be π and standard diversion be π. If π ( πβ€39 ) =0.0548 and π=63, receive π use the criterion convention distribution table .

## Answer

in ordering to determine the nameless standard diversion π, we code π by the exchange of variable πβ¦π=πβππ, where the entail π=63. nowadays πβΌποΉ0,1ο ο¨ surveil the standard normal distribution and π ( πβ€39 ) =ποΌπβ€39β63πο=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 πβ€β1.6 . thus, we have 39β63π=β1.6π=β24β1.6=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 π cost vitamin a random variable that equal normally distributed with base π and standard diversion π. give that π ( πβ€72.44 ) =0.6443 and π ( πβ₯37.76 ) =0.9941, forecast the respect of π and π .

## Answer

in order to discovery the unknown bastardly π and standard deviation π, we code π by the transfer of variable πβ¦π=πβππ. now πβΌποΉ0,1ο ο¨ play along the criterion convention distribution and π ( πβ€72.44 ) =πο½πβ€72.44βππο=0.6443 and π ( πβ₯37.76 ) =πο½πβ₯37.76βππο=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 πβ€0.36998 and that πβ₯β2.51807 . This return the pair of coincident equality 72.44βππ=0.36998 and 37.76βππ=β2.51807. We multiply both equation by π : 72.44βπ=0.36998π,37.76βπ=β2.51807π. then, we subtract the second from the first to catch 72.44βπβ ( 37.76βπ ) =0.36998πβ ( β2.51807π ) 34.68=2.88805π. therefore, we induce π=34.682.88805=12.0081β¦. We can now substitute π=12.0081β¦ back into the equation 72.44βπ=0.36998π, which contribute uracil 72.44βπ=0.36998Γ12.0081β¦π=67.9972β¦. We arrive at rate of π=68 and π=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 πβΌποΉ3.25, πο ο¨. contribute that π ( π > 2π ) =0.1 and π ( π < π ) =0.3, find the value of π and the respect of π. collapse your solution to one decimal place .

## Answer

in club to discover the unknown standard deviation π and constant π, we code π aside the variety of variable πβ¦π=πβππ, where the bastardly be π=3.25. now πβΌποΉ0,1ο ο¨ follow the standard convention distribution and π ( π > 2π ) =ποΌπ > 2πβ3.25πο=0.1 and π ( π < π ) =ποΌπ < πβ3.25πο=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 π > 1.28155 and that π < β0.5244. This yield the pair of coincident equation 2πβ3.25π=1.28155 and πβ3.25π=β0.5244. We breed both equation aside π : 2πβ3.25=1.28155π, πβ3.25=β0.5244π. then, we reproduce the second base of these by two : 2πβ6.5=β1.0488π. We buttocks now rule out π by subtract the moment equation from the first : 2πβ3.25β ( 2πβ6.5 ) =1.28155πβ ( β1.0488π ) 3.25=2.33035ππ=1.39464β¦. To find oneself the value of π, we buttocks utility back into πβ3.25=β0.5244π : πβ3.25=β0.5244Γ1.39464β¦π=3.25β0.7313β¦=2.5186β¦. frankincense, round to one decimal put, we suffer π=1.4 and π=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 and standard deviation twelve curium. give that 10.56 % of the bloom be inadequate than forty-seven curium, determine π .

## Answer

We have a convention random variable star πβΌποΉπ,12ο ο¨ with strange base. To convert the population share of 10.56 % into deoxyadenosine monophosphate probability, we divide by hundred, so we have π ( π < forty-seven ) =0.1056. in orderliness to detect the strange average π, we code π aside the variety of variable πβ¦π=πβππ, where the standard deviation constitute π=12. immediately πβΌποΉ0,1ο ο¨ follow the standard normal distribution and π ( π < forty-seven ) =πο½π < 47βπ12ο=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 π < β1.25027. frankincense, we receive 47βπ12=β1.25027βπ= ( β1.25027 ) Γ12β47π=62.00324, give uracil π=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=63cm and criterion deviation π. 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 πβΌποΉ63, πο ο¨ with nameless variance. To commute the population percentage of 84.13 % into adenine probability, we separate aside hundred, indeed we receive π ( π < seventy-five ) =0.8413. in club to determine the unknown variance πο¨, we code π by the variety of variable πβ¦π=πβππ, where the mean π=63. nowadays πβΌποΉ0,1ο ο¨ comply the standard convention distribution and π ( π < seventy-five ) =ποΌπ < 75β63πο=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 π < 0.99982. therefore, we own 75β63π=0.99982π=120.99982=12.00216β¦. therefore, our variance embody π= ( 12.00216β¦ ) =144ο¨ο¨, to the near integer. lease uracil finish aside recapitulate adenine few crucial concept from this explainer .

## Key Points

- Given a normal random variable πβΌποΉπ,πο ο¨ and a probability π ( π < π₯ ) =πο§ ,we can code π by the change of variables πβ¦π=πβππ,
where πβΌοΉ0,1ο ο¨ .Then, we can use the standard normal distribution to find an unknown mean or standard

deviation.- If we are given two probabilities π(π<π₯)=πο§ and π ( π > π¦ ) =πο¨ ,

then we can derive a pair of simultaneous equations to find the

mean and the standard deviation when both are unknown.