The state bank aims to maintain an average of 1,500 clients with a standard deviation of 850. If a sample of 60 branches is taken, find the probability that the average number of customers is 1,700.$$μ = 1500, \quad σ = 850, \quad P( X = 1700) = {}?$$
\begin{align} P(X = 1700) &= P (X ≤ 1700)-P(X ≥ 1700) \\ &= P (X ≤ 1700) - ( 1 -P (X ≤ 1700)) \\ &= 2P(X ≤ 1700) – 1 \end{align}
P((X -μ) /σ = (1700- 1500/850) ) = 2P( (X -μ) /σ ≤ (1700- 1500)/850) - 1
P (Z = 20/85) = 2P(Z ≤ 20/85)- 1;
P(Z = 0,23) = 2P ( Z ≤ 0,23) – 1
P(x=Z) = 0.5910 P(Z = 0,23) = 2P ( Z ≤ 0,23)- 1 = 2 . 0591 - 1 = 1,182 - 1
= 0,182
= 18,20% P(Z = 0,23)
= 18,20% I'm getting a probability of 18.20% but I'm I right?
I'm sorry, but I don't think you got the right solution. The way you asked the question leads me to think that you want $P(\overline{X} \in [1699.5; 1700.5[)$. And what you solved was $P(X \leq 1700)$, which is different for three reasons:
"Being equal" in a count variable means that you can apply the continuity correction to the computations, so the integer $1700$ would be considered as the interval $[1699.5; 1700.5[$ of real numbers that when rounded give $1700$. Note that if it was asked the probability of the average being at least $1700$ we would solve this problem differently.
Secondly, since you're looking at the average, you should use $\sigma / \sqrt{n}$, where $n=60$ is the number of branches as the standard deviation, because the question is on the average number of clients $\overline{X}$ instead of the number of clients in one agency $X$. In here, I assumed that $\sigma = 850$ was the standard deviation of the number of clients, so $\sigma / \sqrt{n}$ is the standard deviation of the average over $n$ measurements of the number of clients.
Finally $P(X=1700)$ is not equal to $P(X \leq 1700) - P(X \geq 1700)$. If the value was $1400$ you would get a negative probability. The correct formula would be $P(X=1700) = P(X \leq 1700) - P(X < 1700)$. If $X$ was continuous this would give $0$, since it's discrete we can get a non-zero result.