Respuesta :
Answer:
3.24% of rods will be discarded
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 = 25, \sigma = 0.07[/tex]
What proportion of rods will be discarded
Shorter than 24.85.
pvalue of Z when X = 24.85
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{24.85 - 25}{0.07}[/tex]
[tex]Z = -2.14[/tex]
[tex]Z = -2.14[/tex] has a pvalue of 0.0162
Longer than 25.15
1 subtracted by the pvalue of Z when X = 25.15
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{25.15 - 25}{0.07}[/tex]
[tex]Z = 2.14[/tex]
[tex]Z = 2.14[/tex] has a pvalue of 0.9838
1 - 0.9838 = 0.0162
2*0.0162 = 0.0324
3.24% of rods will be discarded
Answer:
The proportion of rods will be discarded is 0.03236
Step-by-step explanation:
From the question;
mean; μ = 25 cm
Standard deviation; σ = 0.07 cm)
Now, this is a normal distribution problem where we are supposed to find the Z-score.
Z is given by;
Z = (X - μ)/0.07
Applying this to the question of, rods that are shorter than 24.85 cm or longer than 25.15 cm are discarded. We arrive at;
P(X<24.85) + P(X>25.15)
This will give us;
P(z < [(24.85 - 25)/0.07)]) + P(z > [(25.15 - 25)/0.07)])
= P(z < (-2.143)) + P(z > 2.14)
This now transforms to;
= P(z ≤ (-2.143)) + 1 - P(z ≤ 2.143)
From standard normal table which i have attached,
P(z ≤ (-2.143)) = 0.01618 and
P(z ≤ 2.143) = 0.98382
Thus,
P(X<24.85) + P(X>25.15) = 0.01618 + 1 - 0.98382 = 0.03236
