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2 answers:
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Answer:
Step-by-step explanation:
Hello!
The variable of interest is the readings on thermometers. This variable is normally distributed with mean μ= 0 degrees C and standard deviation σ= 1.00 degrees C.
The objective is to find the readings that are in the top 3.3% of the distribution and the lowest 3.3% of the distribution.
Symbolically:
The lower value P(X≤a)=0.033
Top value P(X≥b)=0.033
(see attachment)
Lower value:
The accumulated probability until "a" is 0.03, since the variable has a normal distribution, to reach the value of temperature that has the lowest 3.3%, you have to work under the standard normal distribution.
First we look the Z value corresponding to 0.033 of probability:
Z= -1.838
Now you reverste the standardization using the formula Z= (a-μ)/δ
a= (Z*δ)+μ
a= (-1.838*1)+0
a= -1.838
Top value:
P(X≥b)=0.033
This value has 0.033 of the distribution above it then 1 - 0.033= 0.967
is below it.
You can rewrite the expression as:
P(X≤b)=0.967
Now you have to look the value of Z that corresponds to 0.967 of accumulated probability:
b= (Z*δ)+μ
b= (1.838*1)+0
b= 1.838
The cutoff values that separates rejected thermometers from the others are -1.838 and 1.838 degrees C.
I hope it helps!
Answer: a)
Step-by-step explanation:
Given : For safety reasons, four different alarm systems were installed in the vault containing the safety deposit boxes at a Beverly Hills bank.
Each of the four systems detects theft with a probability of .99 independently of the others .
Probability for independent events occurs together =
.
Then, the probability that when a theft occurs, all four systems will detect it
= (0.99) x (0.99) x (0.99) x (0.99)
Hence, the probability that when a theft occurs, all four systems will detect it is .
Hence, the correct answer is a) .