Math  /  Data & Statistics

QuestionAssume that the weights of ripe watermelons grown at a particular farm are normally distributed with a mean of 90 pounds and a standard deviation of 2.5 pounds. Determine the percent of watermelons that weigh between 88.34 pounds and 94.61 pounds. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. \square \% (Round to two decimal places as needed.)

Studdy Solution

STEP 1

What is this asking? What percentage of watermelons weigh between **88.34** pounds and **94.61** pounds, assuming a normal distribution with a mean of **90** pounds and a standard deviation of **2.5** pounds? Watch out! Don't forget to look up the z-scores correctly and use the table carefully!

STEP 2

1. Calculate the z-scores.
2. Look up the probabilities.
3. Calculate the final probability.

STEP 3

Alright, let's **calculate the z-score** for 88.3488.34 pounds!
The z-score tells us how many standard deviations a value is away from the mean.
The formula is z=xμσz = \frac{x - \mu}{\sigma}, where xx is our value, μ\mu is the mean, and σ\sigma is the standard deviation.
Plugging in our values, we get z=88.34902.5z = \frac{88.34 - 90}{2.5}.

STEP 4

This simplifies to z=1.662.5=0.664z = \frac{-1.66}{2.5} = -0.664.
So, **88.34** pounds is **0.664** standard deviations *below* the mean.

STEP 5

Now, let's do the same for **94.61** pounds!
Using the same formula, we have z=94.61902.5z = \frac{94.61 - 90}{2.5}.

STEP 6

This simplifies to z=4.612.5=1.844z = \frac{4.61}{2.5} = 1.844.
So, **94.61** pounds is **1.844** standard deviations *above* the mean.
Awesome!

STEP 7

Using our z-table, we find the probability associated with a z-score of 0.66-0.66 is **0.2454**, and for 0.67-0.67 it's **0.2514**.
Since our z-score is 0.664-0.664, we can estimate the probability to be roughly halfway between these two values, which is approximately **0.2484**.
This means that about **24.84%** of watermelons weigh *less* than **88.34** pounds.

STEP 8

Now, for the z-score of 1.841.84, the table gives us **0.9671**, and for 1.851.85 it's **0.9678**.
Since our z-score is 1.8441.844, we can estimate the probability to be around **0.9675**.
This means about **96.75%** of watermelons weigh *less* than **94.61** pounds.

STEP 9

To find the percentage of watermelons between **88.34** and **94.61** pounds, we subtract the smaller probability from the larger one: 0.96750.2484=0.71910.9675 - 0.2484 = 0.7191.

STEP 10

Multiplying by 100 to get a percentage, we find that approximately **71.91%** of the watermelons weigh between **88.34** and **94.61** pounds.

STEP 11

71.9171.91% of the watermelons weigh between 88.34 pounds and 94.61 pounds.

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