The length of z in the triangle

A certain brand of flood lamps has a lifetime that is normally distributed with a mean of 3,750 hours and a standard deviation o

vesna_86 [32]

Given that mean=3750 hours and standard deviation is 300:
Then:
a. The probability that a lamp will last for more than 4,000 hours?
P(x>4000)=1-P(x<4000)
but
P(x<4000)=P(z<Z)
where:
z=(x-μ)/σ
z=(4000-3750)/300
z=0.833333
thus
P(x<4000)=P(z<0.8333)=0.7967
thus
P(x>4000)=1-0.7967=0.2033

b.What is the probability that a lamp will last less than 3,000 hours?
P(x<3000)=P(z<Z)
Z=(3000-3750)/300
z=-2.5
thus
P(x<3000)=P(z<-2.5)=0.0062

c. .What lifetime should the manufacturer advertise for these lamps in order that only 4% of the lamps will burn out before the advertised lifetime?
the life time will be found as follows:
let the value be x
the value of z corresponding to 0.04 is z=-2.65
thus
using the formula for z-score:
-2.65=(x-3750)/300
solving for x we get:
-750=x-3750
x=-750+3750
x=3000

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I running track is 400 meters. if i make 4 laps, what is the distance that i traveled?

IgorLugansk [536]

You would have ran 1,200 meters

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( BRAINLIEST IF CORRECT)If your business is generating $500 per month in cash flow, and you are trying to save up for a new pitc

svet-max [94.6K]

9 months it gives you $200 extra money

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Need help with the two questions please:)

____ [38]

The decimal form is x=1.2 and the mixed number is x= 1 1/5

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Read 2 more answers

A professor receives, on average, 24.7 emails from students the day before the midterm exam. To compute the probability of recei

MaRussiya [10]

Answer:

Let X the random variable that represent the number of emails from students the day before the midterm exam. For this case the best distribution for the random variable X is $X \sim Poisson(\lambda=24.7)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

The best answer for this case would be:

C. Poisson distribution

Step-by-step explanation:

Let X the random variable that represent the number of emails from students the day before the midterm exam. For this case the best distribution for the random variable X is $X \sim Poisson(\lambda=24.7)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$