Find the following probabilities from z-scores.=NORM.S.DIST(Z 'TRUE')(A) P(z<−1.3)=−−−P\left(z<-1.3\right)=_{---}(B) P(Z>2.4)=−−−P\left(Z>2.4\right)=_{---}(C) P(−1.3≤Z≤2.4)=−−−P\left(-1.3\le Z\le2.4\right)=_{---}

(A) = 0.096800485 0.096800485 0.096800485 ; (B) = 0.008197536 0.008197536 0.008197536 ; (C) = 0.895001979 0.895001979 0.895001979

Find the following probabilities from z-scores. =NORM.S.DIST(Z 'TRUE') (A) P ( z < − 1.3 ) = − − − P\left(z<-1.3\right)=_{---} (B) P ( Z > 2.4 ) = − − − P\left(Z>2.4\right)=_{---} (C) P ( − 1.3 ≤ Z ≤ 2.4 ) = − − − P\left(-1.3\le Z\le2.4\right)=_{---}

Welcome to the practice. In this problem, we'd like to find the following probabilities from z scores. Because we're using z scores, we'll have to use the function equals norm dot s dot dist because we have the standard normal distribution, as these scores have been standardized to fit that distribution. Let's start with part a, finding the probability that c is less than negative 1.3. The equals norm dot s dot dist function finds a probability that c is less than a certain value, so we can find this probability directly. I'm gonna come over here, and I can either copy and paste in my function or type it in. Norm dot s dot dist. Then I need to enter in my arguments. My z score, I can see from the problem, is negative 1.3. And we always wanna enter in true for the second argument because we wanna find the probability that we are less than that z score. I'm gonna click enter, and we get that probability at about 0.1. So we have approximately a point one probability or about a 10% chance of getting a z score less than negative 1.3. Now in part b, we wanna find the probability that c is greater than 2.4. This time, I wanna find a right probability, a probability of getting a z score higher than 2.4. And as a reminder, the norm.s. Dist function finds a left probability, a probability that we are less than a certain z score. So I'm gonna have to take one minus the probability to the left to find my probability to the right. So I'm gonna start with the equal sign to let Excel know I want to run a calculation and then do one minus, and then I wanna type in the formula to find the probability to the left. Again, one minus the probability to the left gives me the probability to the right. I can use the norm dot s dot dist function to find the probability to the left of 2.4. As a reminder, I don't need to retype the equal sign because I already have one at the beginning of my formula. So I'm gonna type norm dot s dot dist open parenthesis. I'm gonna enter in my z score of 2.4, and I'm gonna enter in true for that second argument. So if I look at this formula bar here, I have one minus this norm dot s dot dist term, where that norm dot s dot dist term is giving me the probability that we are less than 2.4. So I'm taking one minus the probability that c is less than 2.4, and I get about 0.008, which is the probability that z is greater than 2.4. Now let's look at part c, where we wanna find the probability that negative 1.3 is less than or equal to z is less than or equal to 2.4, or the probability that z is in between negative one point three and two point four. So I wanna take out the two wings. I wanna take out the wing to the left of negative 1.3 and the wing to the right of 2.4 and find the area in the middle or find that probability in the middle. I actually found the two wings in part a and part b. So I'm gonna take the total probability under the normal curve, which is one, and subtract out the two probabilities I found in a and b. So I'm gonna come over to the cell where I want it. I'm gonna type my equal sign to start because I wanna do a calculation in Excel. Then I'm gonna type one minus, and I'm gonna click on this probability first from part a to subtract out my left tail that I don't want, the probability that z is less than negative 1.3, and then subtract my other tail to the right that I don't want, the probability that z is greater than 2.4. So I have one minus the two tails I don't want, the total area under the normal curve minus the two pieces I don't want. And when I click enter, I get that probability at about 0.9. So that's going to be the probability that we have a z score between negative one point three and two point four. Great work.

Find the following probabilities from z-scores. =NORM.S.DIST(Z 'TRUE') (A) P ( z < − 1.3 ) = − − − P\left(z<-1.3\right)=_{---} (B) P ( Z > 2.4 ) = − − − P\left(Z>2.4\right)=_{---} (C) P ( − 1.3 ≤ Z ≤ 2.4 ) = − − − P\left(-1.3\le Z\le2.4\right)=_{---}

(A) = 0.096800485 0.096800485 0.096800485 ; (B) = 0.008197536 0.008197536 0.008197536 ; (C) = 0.895001979 0.895001979 0.895001979

(A) = 0.903199515 0.903199515 0.903199515 ; (B) = 0.008197536 0.008197536 0.008197536 ; (C) = 0.895001979 0.895001979 0.895001979

(A) = 0.903199515 0.903199515 0.903199515 ; (B) = 0.991802464 0.991802464 0.991802464 (C) = 0.895001979 0.895001979 0.895001979

(A) = 0.096800485 0.096800485 0.096800485 ; (B) = 0.008197536 0.008197536 0.008197536 ; (C) = 1.088602949 1.088602949 1.088602949

6. Normal Distribution & Continuous Random Variables

Finding Probabilities, Z Values, and X Values with the Normal Distribution-Excel

Statistics for Business

6. Normal Distribution & Continuous Random Variables

Finding Probabilities, Z Values, and X Values with the Normal Distribution-Excel

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Multiple Choice

Find the following probabilities from z-scores.
=NORM.S.DIST(Z 'TRUE')
(A) $P (z < - 1.3) =_{- - -}$
(B) $P (Z > 2.4) =_{- - -}$
(C) $P (- 1.3 \leq Z \leq 2.4) =_{- - -}$

(A) = $0.903199515$ ; (B) = $0.008197536$ ; (C) = $0.895001979$

(A) = $0.096800485$ ; (B) = $0.008197536$ ; (C) = $0.895001979$

(A) = $0.903199515$ ; (B) = $0.991802464$ (C) = $0.895001979$

(A) = $0.096800485$ ; (B) = $0.008197536$ ; (C) = $1.088602949$

Verified step by step guidance

Step 1: Understand that the problem involves finding probabilities related to the standard normal distribution, where the variable Z follows a normal distribution with mean 0 and standard deviation 1.

Step 2: For part (A), to find $P(Z < -1.3)$, use the cumulative distribution function (CDF) of the standard normal distribution, which can be found using the function $\text{NORM.S.DIST}(-1.3, \text{TRUE})$. This gives the probability that Z is less than -1.3.

Step 3: For part (B), to find $P(Z > 2.4)$, first find the CDF value at 2.4, i.e., $\text{NORM.S.DIST}(2.4, \text{TRUE})$, which gives $P(Z \leq 2.4)$. Then subtract this value from 1 to get $P(Z > 2.4) = 1 - P(Z \leq 2.4)$.

Step 4: For part (C), to find $P(-1.3 \leq Z \leq 2.4)$, calculate the CDF values at both points: $P(Z \leq 2.4)$ and $P(Z < -1.3)$. Then subtract the smaller CDF value from the larger one: $P(-1.3 \leq Z \leq 2.4) = P(Z \leq 2.4) - P(Z < -1.3)$.

Step 5: Use a standard normal table, calculator, or software function like $\text{NORM.S.DIST}$ in Excel to find the exact CDF values for the given z-scores, then apply the above steps to get the required probabilities.

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