Final answer:
To find the value of c for the interval P(c≤Z≤1.13)=0.8386 in a standard normal distribution, use a z-table to find the cumulative area to the right of c, by subtracting the area to the right of the given z-score of 1.13 from 1 and then using the z-score corresponding to the result.
Step-by-step explanation:
To find the value of c such that P(c≤Z≤1.13)=0.8386, we first consider the entire area under the standard normal curve to be 1. The problem provides the probability of Z being between c and 1.13, hence we need to find the value of c for the left-tail such that the central area is 0.8386.
Using a standard normal probability table or calculator, find the z-score that leaves an area of 0.8386 to the right. If we are given a table that shows the area to the left of Z, we will look up 1 - 0.8386 which is 0.1614. The closest value in the table corresponds to a z-score of approximately -0.99. Therefore, we can estimate c ≈ -0.99.
To verify this, look for the area to the left of Z=1.13, which is approximately 0.8708. Subtracting the left-tail area corresponding to c, which is approximately 0.1614, from this area gives an approximate probability of 0.8708 - 0.1614 = 0.7094. This indicates an error because 0.7094 is not consistent with the given central area of 0.8386.
To correct this, we need to find the area to the left of Z=1.13, which is typically found in a z-table. If it isn't provided, we use the TI-83, 83+, or 84+ calculator command invNorm(0.8708,0,1) to find the z-score corresponding to that area. Knowing the full area under the curve is 1, we subtract from this value to find the area in the left tail, which should lead us to the correct value of c.