Math / Data & Statistics

QuestionThe lengths of mature trout in a local lake are approximately normally distributed with a mean of $\mu=13.7$ inches, and a standard deviation of $\sigma=1.6$ inches.
Fill in the indicated boxes.
Find the z -score corresponding to a fish that is $\mathbf{1 3 . 3}$ inches long. Round your answer to the nearest hundredth as needed. $z=$ $\square$ How long is a fish that has a z -score of 0.2 ? Round your answer to the nearest tenth as needed. $\square$ inches

Studdy Solution

STEP 1

What is this asking? We're looking for the z-score of a 13.3-inch trout and the length of a trout with a z-score of 0.2, given the mean and standard deviation of trout lengths. Watch out! Don't mix up the z-score formula!
It's (value - mean) / standard deviation, not (mean - value) / standard deviation.
Also, pay close attention to rounding rules!

STEP 2

1. Calculate the z-score for a 13.3-inch trout.
2. Calculate the length of a trout with a z-score of 0.2.

STEP 3

Alright, let's dive in!
We're given the length of the fish, which is $13.3$ inches.
We also know the mean length ( $\mu = 13.7$ ) and the standard deviation ( $\sigma = 1.6$ ).

STEP 4

The z-score formula is a magical tool that tells us how far away a value is from the mean, in terms of standard deviations.
It's like a universal translator for data!
The formula is: $z = \frac{x - \mu}{\sigma}$ Where $x$ is our value, $\mu$ is the mean, and $\sigma$ is the standard deviation.

STEP 5

Let's plug in our values: $x = 13.3$ , $\mu = 13.7$ , and $\sigma = 1.6$ . $z = \frac{13.3 - 13.7}{1.6}$

STEP 6

Time to crunch the numbers! $z = \frac{-0.4}{1.6}$ $z = -0.25$ So, the z-score for a 13.3-inch trout is $-0.25$ .
This means the 13.3-inch trout is 0.25 standard deviations below the mean.

STEP 7

Now, we're given the z-score ( $z = 0.2$ ) and need to find the length of the fish.
We still know the mean ( $\mu = 13.7$ ) and standard deviation ( $\sigma = 1.6$ ).

STEP 8

We can rearrange our trusty z-score formula to solve for $x$ : $z = \frac{x - \mu}{\sigma}$ Multiply both sides by $\sigma$ to get one step closer to isolating $x$ : $z \cdot \sigma = x - \mu$ Add $\mu$ to both sides to isolate $x$ : $x = z \cdot \sigma + \mu$

STEP 9

Let's plug in our values: $z = 0.2$ , $\mu = 13.7$ , and $\sigma = 1.6$ . $x = 0.2 \cdot 1.6 + 13.7$

STEP 10

Time for some more number crunching! $x = 0.32 + 13.7$ $x = 14.02$ We need to round to the nearest tenth, so the length of the fish is $14.0$ inches.

STEP 11

A fish that is 13.3 inches long has a z-score of $-0.25$ .
A fish that has a z-score of 0.2 is 14.0 inches long.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the z-score for a 13.3-inch trout.
2. Calculate the length of a trout with a z-score of 0.2.

STEP 3

Alright, let's dive in!
We're given the length of the fish, which is $13.3$ inches.
We also know the mean length ( $\mu = 13.7$ ) and the standard deviation ( $\sigma = 1.6$ ).

STEP 4

The z-score formula is a magical tool that tells us how far away a value is from the mean, in terms of standard deviations.
It's like a universal translator for data!
The formula is: $z = \frac{x - \mu}{\sigma}$ Where $x$ is our value, $\mu$ is the mean, and $\sigma$ is the standard deviation.

STEP 5

Let's plug in our values: $x = 13.3$ , $\mu = 13.7$ , and $\sigma = 1.6$ . $z = \frac{13.3 - 13.7}{1.6}$

STEP 6

Time to crunch the numbers! $z = \frac{-0.4}{1.6}$ $z = -0.25$ So, the z-score for a 13.3-inch trout is $-0.25$ .
This means the 13.3-inch trout is 0.25 standard deviations below the mean.

STEP 7

Now, we're given the z-score ( $z = 0.2$ ) and need to find the length of the fish.
We still know the mean ( $\mu = 13.7$ ) and standard deviation ( $\sigma = 1.6$ ).

STEP 8

We can rearrange our trusty z-score formula to solve for $x$ : $z = \frac{x - \mu}{\sigma}$ Multiply both sides by $\sigma$ to get one step closer to isolating $x$ : $z \cdot \sigma = x - \mu$ Add $\mu$ to both sides to isolate $x$ : $x = z \cdot \sigma + \mu$

STEP 9

Let's plug in our values: $z = 0.2$ , $\mu = 13.7$ , and $\sigma = 1.6$ . $x = 0.2 \cdot 1.6 + 13.7$

STEP 10

Time for some more number crunching! $x = 0.32 + 13.7$ $x = 14.02$ We need to round to the nearest tenth, so the length of the fish is $14.0$ inches.

STEP 11

A fish that is 13.3 inches long has a z-score of $-0.25$ .
A fish that has a z-score of 0.2 is 14.0 inches long.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the *z*-score for a 13.3-inch trout.2. Calculate the length of a trout with a *z*-score of 0.2.

STEP 3

Alright, let's **dive in**!We're given the length of the fish, which is 13.313.313.3 inches.We also know the **mean** length (μ=13.7\mu = 13.7μ=13.7) and the **standard deviation** (σ=1.6\sigma = 1.6σ=1.6).

STEP 4

STEP 5

Let's **plug in** our values: x=13.3x = 13.3x=13.3, μ=13.7\mu = 13.7μ=13.7, and σ=1.6\sigma = 1.6σ=1.6. z=13.3−13.71.6z = \frac{13.3 - 13.7}{1.6}z=1.613.3−13.7​

STEP 6

Time to **crunch the numbers**! z=−0.41.6z = \frac{-0.4}{1.6}z=1.6−0.4​ z=−0.25z = -0.25z=−0.25So, the *z*-score for a 13.3-inch trout is −0.25-0.25−0.25.This means the 13.3-inch trout is 0.25 standard deviations *below* the mean.

STEP 7

Now, we're given the *z*-score (z=0.2z = 0.2z=0.2) and need to find the length of the fish.We still know the **mean** (μ=13.7\mu = 13.7μ=13.7) and **standard deviation** (σ=1.6\sigma = 1.6σ=1.6).

STEP 8

STEP 9

Let's **plug in** our values: z=0.2z = 0.2z=0.2, μ=13.7\mu = 13.7μ=13.7, and σ=1.6\sigma = 1.6σ=1.6. x=0.2⋅1.6+13.7x = 0.2 \cdot 1.6 + 13.7x=0.2⋅1.6+13.7

STEP 10

Time for some more **number crunching**! x=0.32+13.7x = 0.32 + 13.7x=0.32+13.7 x=14.02x = 14.02x=14.02We need to round to the nearest tenth, so the length of the fish is 14.014.014.0 inches.

STEP 11

A fish that is 13.3 inches long has a *z*-score of −0.25-0.25−0.25.A fish that has a *z*-score of 0.2 is 14.0 inches long.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Calculate the z-score for a 13.3-inch trout.
2. Calculate the length of a trout with a z-score of 0.2.

Alright, let's dive in!
We're given the length of the fish, which is $13.3$ inches.
We also know the mean length ( $\mu = 13.7$ ) and the standard deviation ( $\sigma = 1.6$ ).

Let's plug in our values: $x = 13.3$ , $\mu = 13.7$ , and $\sigma = 1.6$ . $z = \frac{13.3 - 13.7}{1.6}$

Time to crunch the numbers! $z = \frac{-0.4}{1.6}$ $z = -0.25$ So, the z-score for a 13.3-inch trout is $-0.25$ .
This means the 13.3-inch trout is 0.25 standard deviations below the mean.

Now, we're given the z-score ( $z = 0.2$ ) and need to find the length of the fish.
We still know the mean ( $\mu = 13.7$ ) and standard deviation ( $\sigma = 1.6$ ).

Let's plug in our values: $z = 0.2$ , $\mu = 13.7$ , and $\sigma = 1.6$ . $x = 0.2 \cdot 1.6 + 13.7$

Time for some more number crunching! $x = 0.32 + 13.7$ $x = 14.02$ We need to round to the nearest tenth, so the length of the fish is $14.0$ inches.

A fish that is 13.3 inches long has a z-score of $-0.25$ .
A fish that has a z-score of 0.2 is 14.0 inches long.