Specifications for an aircraft bolt require that the ultimate tensile strength be at least 18 kn. it is known that 10% of the bolts have strengths less than 18.3 kn and that 5% of the bolts have strengths greater than 19.76 kn. it is also known that the strengths of these bolts are normally distributed.

a. find the mean and standard deviation of the strengths

b. what proportion of the bolts meet the strength specification?

Mean strength is 18.94 and your standard deviation is 0.5.

So the proportion of bolts that meet the specifications is 97%.

Step-by-step explanation:

First of all determine the z- scores of these points.There are 10% of bolts with a strength less than 18.3 kN and this normally distributed you can use chart or calculator to calculate z-score. As i have 5% then z-score is -1.28.Then check the other 19.76kN then find that it has a z-score of 1.64.

To check the difference subtract 19.76 and 18.3 then you get 1.46.

Subtract z-scores 1.64 - (-1.28) = 2.92

Then standard deviation is 1.46/2.92 = 0.50

mean of the bolts is obtained by adding 1.28 *0.5 = 0.64 to 18.3 then

subtract 1.64 *0.5 = 0.82 to 19.76

Then mean is 18.94

Mean strength is 18.94 and your standard deviation is 0.5.

For strength specification. First, we find the z-score for this value:

(18-18.94)/0.5=-1.88

the probability of a bolt being made stronger than this z-score.It is approximately 0.97.

So the proportion of bolts that meet the specifications is 97%. .

50-38=12

step-by-step explanation:

b.my answer is b.

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