Answer :
Answer:
2%
Step-by-step explanation:
We are given;
Standard Design mean; μ = 18
Sample design mean; x¯ = 15
Standard Standaed deviation; σ = 1.4
Let's find the test statistic from the formula;
z = (x¯ - μ)/σ
Thus;
z = (15 - 18)/1.4
z = -2.14
From the z-table attached, the p-value at z = -2.14 is 0.01618
Thus,
P(X < 15) = 0.01618
Converting to percentage, we have
P(X < 15) = 1.618%
This is approximately 2%
Using the normal distribution, it is found that there is a 2% probability that a fuel injection system will last less than 15 years.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of 18 years, hence [tex]\mu = 18[/tex].
- The standard deviation is of 1.4 years, hence [tex]\sigma = 1.4[/tex].
The probability that a fuel injection system will last less than 15 years is 1 subtracted by the p-value of Z when X = 15, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{15 - 18}{1.4}[/tex]
[tex]Z = -2.14[/tex]
[tex]Z = -2.14[/tex] has a p-value of 0.02.
0.02 = 2% probability that a fuel injection system will last less than 15 years.
To learn more about the normal distribution, you can take a look at https://brainly.com/question/24663213