Respuesta :
Answer:
1)
So 4.75% probability of a snowboard being defective
2)
3.22% probability that the number of defective boards is greater than 10
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X
In this problem, we have that:
[tex]\mu = 10, \simga = 3[/tex]
1. What’s the probability of a snowboard being defective?
This is the pvalue of Z when X = 5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 10}{3}[/tex]
[tex]Z = -1.67[/tex]
[tex]Z = -1.67[/tex] has a pvalue of 0.0475.
So 4.75% probability of a snowboard being defective
2. In a shipment of 120 snowboards, what is the probability that the number of defective boards is greater than 10?
We use the binomial approximation to the normal to solve this question.
aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Can be approximated with mean [tex]\mu = E(X) = np[/tex] and standard deviation [tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
4.75% probability of a snowboard being defective, which means that [tex]p = 0.0475[/tex]
So
[tex]\mu = E(X) = np = 120*0.0475 = 5.7[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.0475*0.9525} = 2.33[/tex]
The probability is 1 subtracted by the pvalue of Z when X = 10. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10 - 5.7}{2.33}[/tex]
[tex]Z = 1.85[/tex]
[tex]Z = 1.85[/tex] has a pvalue of 0.9678
1 - 0.9678 = 0.0322
3.22% probability that the number of defective boards is greater than 10
1) The probability that a snowboard is defective is; 4.746%
2) The probability that the number of defective boards is greater than 10 is; 3.216%
1) We are told that the average life of their product is 10 years. Thus;
Mean; μ = 10 years
A snowboard is considered defective if its life is less than 5 years. Thus;
x' = 5
standard deviation for the life of a board is 3 years. Thus; σ = 3
To find the probability that a snowboard is defective means;
P(X = 5).
Let us first find the z-score.
z = (x' - μ)/σ
z = (5 - 10)/3
z = -1.67
From online p-value from z-score calculator at z = -1.67;
P(X = 5) = 0.04746 or 4.746%
2) We are told that in a shipment of 120 snowboards, the number that are defective is greater than 10.
Thus; n = 120
From expected value theorem, we know that;
E(X) = μ = np
μ = (120 × 0.04746)
μ = 5.6952
Now, the standard deviation is given by;
σ = √(np(1 - p))
σ = √[(120 × 0.04746)(1 - 0.04746)]
σ = 2.33
Thus for P(X > 10), let us first find the z-score;
z = (10 - 5.6952)/2.33
z = 1.85
From online p-value from z-score calculator at z = 1.85;
P(X > 10) = 0.03216 or 3.216%
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