The mass of a species of mouse commonly found in houses is normally distributed with a mean of 19.2
grams with a standard deviation of 0.15 grams.
Use Stapplet or another Normal Calculator to find the probabilities and cutoff values. For parts (a)
through ©, enter a decimal with 4 decimal places. Use 1 place for part .
a) What is the probability that a randomly chosen mouse has a mass of less than 19.14 grams?
b) What is the probability that a randomly chosen mouse has a mass of more than 19.37 grams?
c) What proportion of mice have a mass between 19.11 and 19.32 grams?
d) of all mice have a mass of less than

Question

The mass of a species of mouse commonly found in houses is normally distributed with a mean of 19.2
grams with a standard deviation of 0.15 grams.
Use Stapplet or another Normal Calculator to find the probabilities and cutoff values. For parts (a)
through ©, enter a decimal with 4 decimal places. Use 1 place for part

.
a) What is the probability that a randomly chosen mouse has a mass of less than 19.14 grams?
b) What is the probability that a randomly chosen mouse has a mass of more than 19.37 grams?
c) What proportion of mice have a mass between 19.11 and 19.32 grams?
d)

of all mice have a mass of less than

UpStudy AI · Accepted Answer

To solve these problems, we will use the properties of the normal distribution. Given that the mass of the mice is normally distributed with a mean ($$\mu$$) of 19.2 grams and a standard deviation ($$\sigma$$) of 0.15 grams, we can standardize the values using the z-score formula:

$$
z = \frac{(X - \mu)}{\sigma}
$$

where $$X$$ is the value we are interested in.

### a) Probability that a randomly chosen mouse has a mass of less than 19.14 grams

1. Calculate the z-score for $$X = 19.14$$:
   $$
   z = \frac{(19.14 - 19.2)}{0.15} = \frac{-0.06}{0.15} = -0.4
   $$

2. Use the z-table or a calculator to find the probability corresponding to $$z = -0.4$$:
   $$
   P(Z < -0.4) \approx 0.3446
   $$

So, the probability that a randomly chosen mouse has a mass of less than 19.14 grams is **0.3446**.

### b) Probability that a randomly chosen mouse has a mass of more than 19.37 grams

1. Calculate the z-score for $$X = 19.37$$:
   $$
   z = \frac{(19.37 - 19.2)}{0.15} = \frac{0.17}{0.15} \approx 1.1333
   $$

2. Find the probability corresponding to $$z = 1.1333$$:
   $$
   P(Z < 1.1333) \approx 0.8716
   $$

3. To find the probability of more than 19.37 grams:
   $$
   P(Z > 1.1333) = 1 - P(Z < 1.1333) \approx 1 - 0.8716 = 0.1284
   $$

So, the probability that a randomly chosen mouse has a mass of more than 19.37 grams is **0.1284**.

### c) Proportion of mice with a mass between 19.11 and 19.32 grams

1. Calculate the z-scores for $$X = 19.11$$ and $$X = 19.32$$:
   - For $$X = 19.11$$:
     $$
     z = \frac{(19.11 - 19.2)}{0.15} = \frac{-0.09}{0.15} = -0.6
     $$
   - For $$X = 19.32$$:
     $$
     z = \frac{(19.32 - 19.2)}{0.15} = \frac{0.12}{0.15} = 0.8
     $$

2. Find the probabilities:
   - For $$z = -0.6$$:
     $$
     P(Z < -0.6) \approx 0.2743
     $$
   - For $$z = 0.8$$:
     $$
     P(Z < 0.8) \approx 0.7881
     $$

3. The proportion of mice with a mass between 19.11 and 19.32 grams:
   $$
   P(19.11 < X < 19.32) = P(Z < 0.8) - P(Z < -0.6) \approx 0.7881 - 0.2743 = 0.5138
   $$

So, the proportion of mice with a mass between 19.11 and 19.32 grams is **0.5138**.

### d) 30% of all mice have a mass of less than

To find the cutoff value for which 30% of mice have a mass less than that value, we need to find the z-score that corresponds to the 30th percentile.

1. From the z-table, the z-score that corresponds to 0.30 is approximately $$-0.524$$.

2. Now, we can convert the z-score back to the original mass using the formula:
   $$
   X = \mu + z \cdot \sigma
   $$
   $$
   X = 19.2 + (-0.524) \cdot 0.15 \approx 19.2 - 0.0786 \approx 19.1214
   $$

So, 30% of all mice have a mass of less than **19.1** (rounded to one decimal place).

### Summary of Answers:
a) 0.3446  
b) 0.1284  
c) 0.5138  
d) 19.1
a) 0.3446 b) 0.1284 c) 0.5138 d) 19.1

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