A certain type of thread is manufactured with a mean tensile strength of 76.7 kg and a standard deviation 5.9.?

Assume a sample of 100 items is obtained.
a) do we know the distribution of the tensile strength of any produced item?
b) what is the mean and standard deviation of the sample average?
c)What can you say about the distribution of the sample average?
d) Given your answer in part (c), find the probability that a sample of 100 items will have an average
i) greater than 78.
ii) less than 75.
iii) between 75.8 and 76.8

I don't understand the relationship between "thread" and "items," so I can't answer part (a).

(b)
If I assume that the mean and standard deviation in the headline apply to the "items" in the sample, then the mean of the sampling distribution will also be 76.7 kg, but the standard deviation of the distribution of samples of 100 will be (5.9 kg)/sqrt(100) = 0.59 kg.

(c) If the population was normally distributed, the distribution of the means of samples of 100 will also be normally distributed.

(d) 78 corresponds to z = (78 - 76.7)/0.59 = 2.203
75 corresponds to z = (75 - 76.7) / 0.59 = -2.881
75.8 corresponds to z = -1.525
76.8 corresponds to z = 0.1695

Now go to a normal distribution table.
z-score.....P(z < this z-score)
2.881.....0.9980
2.203.....0.9862
1.525.....0.9364
0.1695.....0.5673

Thus, the answers for (d) are
(i) 1 - 0.9862 = 1.38%
(ii) 99.8%
(iii) 0.5673 - (1 - 0.9364) = 50.37%