Answer:
a) P=0.019
b) P=0.263
c) P=0.794
Step-by-step explanation:
We assume that the poll gives the population's proportion of respondents said that their jobs were sometimes or always stressful (p=0.8).
Then, a sample of size n=190 is taken.
The sample mean is:

The sample standard deviation is:

The probability that 140 or fewer workers find their jobs stressful is:

Note: a correction for continuity is applied.
The probability that more than 155 workers find their jobs stressful is

The probability that the number of workers who find their jobs stressful is between 145 and 158 inclusive is:

Your answer is C.
Hope this helped.
The opposite of the fraction one third would be negative one third. To be opposite, they must have differing signs. One number should be positive and the other number should be negative. It is different from reciprocal. To be a reciprocal, <span>one number should be the flipped fraction, or upside down version, of the other number.</span>
Ben would receive £390 while Alex would only receive £130