PLS HELP ME 16 and 17
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Answer:
Step-by-step explanation:
Hello!
The variable is
X: lifespan of an item
This variable has a normal distribution X~N(μ;σ²)
μ= 5.8 years
σ= 1.7 years
The normal distribution is symmetric, centered on μ, asymptotic to -∞ and +∞. The shortest lifespans will be on the left tail of the distribution and the longest lifespans will be on the right tail. You have to determine how many years the 1% of the items with the shortest lifespan will, this is, the value that separates the bottom 1% of the distribution, symbolically:
P(X≤x₀)= 0.01
Where x₀ represents the value of lifespan you were asked for.
To determine this value you have to work using the standard normal distribution. Derived from the normality distribution is the standard normal distribution. Considering a random variable X with normal distribution, mean μ and variance δ², the variable Z =(X-μ)/δ ~N(0;1) is determined.
The standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation.
So to calculate each of the asked probabilities, you have to first, "transform" the value of the variable to a value of the standard normal distribution Z, then you use the standard normal tables to reach the corresponding probability.
So under the standard normal distribution you have to determine the Z value that divides the bottom 1% of the distribution from the top 99%
P(Z≤z₀)= 0.01
Using the standard normal table you have to determine the value of z₀ (it is located in the left tail of the distribution so the value is negative)
z₀= -2.33
Now you have to "translate" the value of Z to a value of the variable of interest:
z₀ =( x₀-μ)/δ
(z₀*δ) =x₀-μ
(z₀*δ)+μ=x₀
x₀= (-2.33*1.7)+5.8= 1.839 years
The lifespan that marks the shortest 1% of the distribution is 1.839
I hope this helps!
