period 4 assignment:
problem 1: babies sleep an average of 14 hours per day with a standard deviation of 2 hours. consider a group of 56 babies.
a. what percent of babies will sleep more than 16 hours?
b. how many will sleep more than 16 hours?
c. what percent of babies sleep less than 10 hours?
d. how many babies will sleep less than 10 hours?
e. what percent of babies will sleep between 14 and 18 hours?
f. what percent of babies will sleep less than 11 hours? (hint: find the z - score and use a z - table)
g. how many babies of sleep less than 11 hours?

Question

Accepted Answer

To solve these problems, we use the properties of the normal distribution (bell curve) with mean $\mu = 14$ hours and standard deviation $\sigma = 2$ hours. The total percentage under the curve is 100%, and we can use the given percentages in the bell - curve diagram (0.15%, 2.35%, 13.5%, 34%) to find the required percentages and then multiply by the total number of babies (56) to get the number of babies.

### Part (a): What percent of babies will sleep more than 16 hours?
#### Step 1: Find the z - score
The z - score formula is $z=\frac{x-\mu}{\sigma}$. For $x = 16$, $\mu=14$ and $\sigma = 2$. Then $z=\frac{16 - 14}{2}=\frac{2}{2}=1$.
In a normal distribution, the percentage of data to the right of $z = 1$ (more than 16 hours) can be found using the bell - curve percentages. The percentage of data to the right of $z = 1$ is $13.5\%+2.35\% + 0.15\%=16\%$ (or we can use the fact that the area to the left of $z = 1$ is approximately 84%, so the area to the right is $100\% - 84\%=16\%$).
#### Answer: 16%

### Part (b): How many will sleep more than 16 hours?
#### Step 1: Use the percentage from part (a)
We know that the percentage of babies who sleep more than 16 hours is 16% (or 0.16 in decimal form).
#### Step 2: Calculate the number of babies
The number of babies is given by the formula: Number of babies=Total number of babies×Percentage. So, $56	imes0.16 = 8.96\approx9$ (we can also use the exact percentage breakdown: The percentage of data with $z>1$ is $2.35\%+0.15\% + 13.5\%=16\%$. $56	imes0.16=8.96\approx9$).
#### Answer: 9

### Part (c): What percent of babies sleep less than 10 hours?
#### Step 1: Find the z - score
For $x = 10$, $z=\frac{10 - 14}{2}=\frac{- 4}{2}=-2$.
In the normal distribution, the percentage of data to the left of $z=-2$ (less than 10 hours) is $2.35\%+0.15\% = 2.5\%$ (from the bell - curve: the area to the left of $z=-2$ is composed of the 2.35% and 0.15% regions).
#### Answer: 2.5%

### Part (d): How many babies will sleep less than 10 hours?
#### Step 1: Use the percentage from part (c)
The percentage of babies who sleep less than 10 hours is 2.5% or 0.025 in decimal form.
#### Step 2: Calculate the number of babies
Number of babies = $56	imes0.025=1.4\approx1$ (we can also calculate using the exact percentages: The percentage of data with $z < - 2$ is $2.35\%+0.15\%=2.5\%$. $56	imes0.025 = 1.4\approx1$).
#### Answer: 1

### Part (e): What percent of babies will sleep between 14 and 18 hours?
#### Step 1: Find the z - scores
For $x = 14$, $z=\frac{14 - 14}{2}=0$. For $x = 18$, $z=\frac{18 - 14}{2}=\frac{4}{2}=2$.
The percentage of data between $z = 0$ and $z = 2$: The percentage of data between $z = 0$ and $z = 1$ is 34%, between $z = 1$ and $z = 2$ is 13.5%. So the total percentage is $34\%+13.5\%=47.5\%$.
#### Answer: 47.5%

### Part (f): What percent of babies will sleep less than 11 hours?
#### Step 1: Find the z - score
For $x = 11$, $z=\frac{11 - 14}{2}=\frac{-3}{2}=-1.5$.
We can use the bell - curve percentages. The percentage of data to the left of $z=-1.5$: The percentage of data with $z < - 2$ is $2.5\%$ (from part c) and between $z=-2$ and $z=-1$ is 13.5%. The z - score of - 1.5 is in the middle of $z=-2$ and $z=-1$. The percentage of data between $z=-2$ and $z=-1.5$: The percentage between $z=-2$ and $z=-1$ is 13.5%, so the percentage between $z=-2$ and $z=-1.5$ is half of 13.5% (since the normal distribution is symmetric around 0) plus the percentage less than $z=-2$. Wait, a better way:
The percentage of data less than $z=-1.5$: The area to the left of $z=-1.5$ can be calculated as follows. The area to the left of $z=-2$ is 2.5%, the area between $z=-2$ and $z=-1$ is 13.5%. The area between $z=-1.5$ and $z=-1$ is half of the area between $z=-2$ and $z=-1$ (because of symmetry). The area between $z=-2$ and $z=-1$ is 13.5%, so the area between $z=-1.5$ and $z=-1$ is $\frac{13.5\%}{2}=6.75\%$. So the area to the left of $z=-1.5$ is $2.5\%+6.75\% = 9.25\%$? Wait, no. Let's use the standard normal table. The cumulative probability for $z=-1.5$ is $P(Z < - 1.5)=0.0668$ or 6.68% $\approx6.7\%$. But from the given bell - curve (with 0.15%, 2.35%, 13.5%, 34%), the percentage of data less than $z=-1$ is $13.5\%+2.35\%+0.15\% = 16\%$, less than $z=-2$ is $2.35\%+0.15\%=2.5\%$. The z - score of - 1.5: The percentage of data between $z=-2$ and $z=-1$ is 13.5%, so the percentage of data between $z=-2$ and $z=-1.5$ is $\frac{13.5\%}{2}=6.75\%$. So the percentage of data less than $z=-1.5$ is $2.5\%+6.75\%=9.25\%$. But if we use the z - table, $P(Z < - 1.5)=0.0668\approx6.7\%$. There is a discrepancy because the given bell - curve is a simplified version (using approximate percentages for integer z - scores). Let's go back to the problem's hint (use z - table). The z - score for $x = 11$ is $z=\frac{11 - 14}{2}=-1.5$. Looking up $z=-1.5$ in the standard normal table, the cumulative probability $P(Z < - 1.5)=0.0668\approx6.7\%$. But if we use the given percentages in the diagram:
The percentage of data less than 10 hours (z=-2) is 2.5%, between 10 and 12 hours (z=-2 to z=-1) is 13.5%, between 12 and 14 hours (z=-1 to z = 0) is 34%. Wait, no, the original mean is 14, so 14 is the center. 14 - 2=12, 14 - 4 = 10, 14+2 = 16, 14 + 4=18.
The percentage of data less than 11 hours: 11 is 14 - 3, so it's 0.15% (less than 10 - 2=8? No, the diagram has 0.15% at the far left, 2.35% next, 13.5%, 34%. Wait, maybe the diagram is for z=-3, z=-2, z=-1, z = 0, z = 1, z = 2, z = 3.
The percentage of data less than $z=-1.5$: The percentage of data with $z < - 2$ is 0.15%+2.35%=2.5%, and between $z=-2$ and $z=-1$ is 13.5%. The z - score of - 1.5 is in the middle of $z=-2$ and $z=-1$. So the percentage of data between $z=-2$ and $z=-1.5$ is half of 13.5% (6.75%). So the total percentage less than $z=-1.5$ is $2.5\%+6.75\%=9.25\%$. But using the z - table, it's approximately 6.7%. However, since the problem gives a hint to use the z - table, we'll go with the z - table value. The z - score for $x = 11$ is $z=\frac{11 - 14}{2}=-1.5$. From the standard normal table, $P(Z < - 1.5)=0.0668\approx6.7\%$.
#### Answer: Approximately 6.7% (or using the diagram - based approach, approximately 9.25%)

### Part (g): How many babies of sleep less than 11 hours?
#### Step 1: Use the percentage from part (f)
If we take the percentage from the z - table (6.7% or 0.067) or from the diagram - based approach (9.25% or 0.0925).
#### Step 2: Calculate the number of babies
If we use the z - table percentage (6.7% or 0.067): $56	imes0.067\approx3.75\approx4$.
If we use the diagram - based percentage (9.25% or 0.0925): $56	imes0.0925 = 5.18\approx5$.
But using the z - score of - 1.5 and the z - table value of 0.0668: $56	imes0.0668\approx3.74\approx4$.
#### Answer: Approximately 4 (or 5 depending on the approach)

### Note
The key idea is to use the normal distribution properties, z - scores, and the given bell - curve percentages (or z - table) to find the required percentages and then multiply by the total number of babies to get the number of babies in each category.

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QUESTION IMAGE

Question

Explanation:

Part (a): What percent of babies will sleep more than 16 hours?

Step 1: Find the z - score

Answer:

Part (b): How many will sleep more than 16 hours?

Step 1: Use the percentage from part (a)

Step 2: Calculate the number of babies