Answer :
The sketch of the normal curve has been attached. A proportion of 38.6% of Nylon ropes will break with a load of 5000 points and with a load of 5056.13 lbs, 95% of Nylon ropes will break.
a)
It is already given that the mean of the distribution is 5036lbs and the standard deviation is 122 lbs.
Hence to sketch the distribution we will take up the apex of the bell shape at the mean i.e 5035 lbs.
A normal distribution usually ranges from 6 to 7 standard deviations.
Therefore, we will take up three standard deviations from the mean on both sides and make the curve.
Hence we get the image attached
b)
Here we need to find the proportion of ropes that will break with a 5000 lbs load.
Hence we need to calculate
P(X<5000)
To do this we will first convert it to a standard normal distribution value
To do that we will use the formula
(x - μ)/σ
(500- - 5036)/ 122
= -36/122
= - 0.295
Now
Φ(-0.295) = 1 - Φ(0.295)
= 1 - 0.614
= 0.386
Therefore the proportion is 38.6%
c)
Here we are asked to calculate the load at which 95% of the ropes will break. Hence we need to calculate a value of x for which the probability will be 0.95
Hence we get
Φ((x - μ)/σ) = 0.95
Φ((x - 5036)/122) = 0.95
Looking up the p-value we find that the z-score of 1.65 has the value closest to it hence we get
Φ((x - 5036)/122) = Φ(1.65)
or, x - 5036 = 1.65 X 122
or, x = 20.13 + 5036
or, x = 5056.13
Complete Question
The climbing rope will break if pulled hard enough. Experiments show that 10.5mm Dynamic nylon rope has a mean breaking point of 5036 lbs with a standard deviation of 122 lbs. Assume breaking points of rope are normally distributed.
a) Sketch the distribution of breaking points for this rope.
b) What proportion of ropes will break with 5000 lbs of load?
c) At what load will 95% of all ropes break?
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