The length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months and standard deviat
ion of 2 months. Find the probability that an instrument produced by this machine will last Q.1. Between 7 and 12 months. Q.2. . Less than 7 months Q.3. More than 5 months but less than 10 months
2 answers:
Answer:
Q.1. 0.4938
Q.2. 0.0062
Q.3. 0.1587
Step-by-step explanation:
Data
- mean, μ = 12
- standard deviation, σ = 2
In the figure attached, the standard normal distribution table is shown.
Z = (x - μ)/σ
Q.1.
if x = 7, then Z = (7 - 12)/2 = -2.5
if x = 12, then Z = (12 - 12)/2 = 0
We want to find:
P(-2.5 < Z < 0) = P(Z < 0) - P(Z < -2.5) = 0.5 - 0.0062 = 0.4938
Q.2.
We want to find:
P(Z < -2.5) = 0.0062
Q.3.
if x = 5, then Z = (5 - 12)/2 = -3.5
if x = 10, then Z = (10 - 12)/2 = -1
We want to find:
P(-3.5 < Z < -1) = P(Z < -1) - P(Z < -3.5) = 0.1587 - 0 = 0.1587
You might be interested in
Answer:
(a) or
(b) or
Step-by-step explanation:
Given
--- North Dakota
--- Cheyenne, Wyoming
Solving (a): Inequality to compare both temperatures
From the given temperatures, we can conclude that:
or
Because
i.e. 1 is greater than -2
or
i.e. -2 is less than 1
Solving (b): Inequality to compare the absolute values of the temperatures
We have:
The absolute values are:
By comparison:
or
Because
i.e. 2 is greater than 1
or
i.e. 1 is less than 2