Find the following z-scores.=NORM.S.INV(P(Z<z))(A) There is 29%29\% chance of a z-score being below what value?z=−−−z=_{---}(B) There is a 67%67\% chance that a z-score is larger than what value?z=−−−z=_{---}

(A) z = − 0.55 z=-0.55 ; (B) z = − 0.44 z=-0.44 z = − 0.44

Find the following z-scores. =NORM.S.INV(P(Z<z)) (A) There is 29 % 29\% chance of a z-score being below what value? z = − − − z=_{---} (B) There is a 67 % 67\% chance that a z-score is larger than what value? z = − − − z=_{---}

Welcome to the practice. In this problem, we'd like to find the following z scores. And because we'll be finding z scores from probabilities, we'll be using the standard normal distribution and the equals norm dot s dot I n v function. Let's start with part a. It says there is a 29% chance of a z score being below what value. So my first step is either going to be to copy and paste in my function or type it in here. I'm gonna type it in equals norm dot s dot I n b, open parenthesis, but it's totally up to you. Now I need to put in my probability. And as a reminder, I need it to be a left hill probability, so I'm going to have to check that first. I can see that this is a 29% chance that a z score is below a certain value. Because it says below, I know that this is a left tail probability, and I can enter in this probability as is. I can either keep it as a percent, 29%, and I will need the percent symbol, or I can convert it to a decimal if that's easier for me. Either way, now that I have my input, I can click enter, and I get that z score at negative 0.55. So I could say that there is a 29% chance of a z score being below negative 0.55. Let's look at part b. It says there is a 67% chance that a z score is larger than what value. So I'm still going to be using my function here because I wanna convert a probability into a z score. And, again, I can either copy and paste it in or type it in, equals norm dot s dot I n v open parenthesis. But this time, I have here a 67% chance that the z score is larger than a certain value. So here, I've been given a right tail probability, and I need to turn it into a left tail probability so I can use it as an input for this function. As a reminder, we said in the concept video that you can take one minus a right tail probability to turn it into a left tail probability. But to do that, I am gonna have to convert my probability here into a decimal, so I'll think of it as 0.67, and I'll have my input be one minus that right tail probability of 0.67. Again, it doesn't matter that my input is an expression. That is totally fine. Now that I have my input taken care of, I can click enter, and I get a z score of negative 0.44. So I can say there is a 67% chance that the z score is larger than negative 0.44. Great job.

Find the following z-scores. =NORM.S.INV(P(Z<z)) (A) There is 29 % 29\% chance of a z-score being below what value? z = − − − z=_{---} (B) There is a 67 % 67\% chance that a z-score is larger than what value? z = − − − z=_{---}

(A) z = − 0.55 z=-0.55 ; (B) z = − 0.44 z=-0.44 z = − 0.44

(A) z = − 0.55 z=-0.55 ; (B) z = 0.44 z=0.44

(A) z = 0.55 z=0.55 ; (B) z = − 0.44 z=-0.44

(A) z = 1.9 z=1.9 ; (B) z = − 0.44 z=-0.44 z = − 0.44

6. Normal Distribution and Continuous Random Variables

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

Statistics

6. Normal Distribution and Continuous Random Variables

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

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Multiple Choice

Find the following z-scores.
=NORM.S.INV(P(Z<z))
(A) There is $29 percent sign$ chance of a z-score being below what value?
$z =_{- - -}$
(B) There is a $67 percent sign$ chance that a z-score is larger than what value?
$z =_{- - -}$

(A) $z = - 0.55$ ; (B) $z equals 0.44$

(A) $z equals 0.55$ ; (B) $z = - 0.44$

(A) $z = - 0.55$ ; (B) $z=-0.44$

(A) $z equals 1.9$ ; (B) $z=-0.44$

Verified step by step guidance

Understand that the problem asks for z-scores corresponding to given probabilities using the standard normal distribution, where the mean is 0 and the standard deviation is 1.

For part (A), identify that the probability given is P(Z < z) = 0.29, meaning the z-score corresponds to the 29th percentile of the standard normal distribution.

Use the inverse standard normal cumulative distribution function, denoted as $\text{NORM.S.INV}(p)$, to find the z-score for a given cumulative probability $p$. For part (A), calculate $z = \text{NORM.S.INV}(0.29)$.

For part (B), note that the probability given is P(Z > z) = 0.67. Since the inverse function uses cumulative probability from the left, convert this to P(Z < z) = 1 - 0.67 = 0.33, then calculate $z = \text{NORM.S.INV}(0.33)$.

Interpret the results: the z-score found in part (A) will be negative because 0.29 is less than 0.5 (the mean), and the z-score in part (B) will also be negative since 0.33 is less than 0.5. This matches the properties of the standard normal distribution.

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