Question
According to a website, the mean number of apps on a smartphone in the United States is 80. Assume the number of apps is normally distributed with mean 80 and standard deviation 20.
(a) What proportion of phones have between 45 and 83 apps? Round the answer to four decimal places.
(a) What proportion of phones have between 45 and 83 apps? Round the answer to four decimal places.
Answers
GPT-4o mini
To find the proportion of phones that have between 45 and 83 apps, we will use the properties of the normal distribution. Given that the number of apps is normally distributed with a mean (\(\mu\)) of 80 and a standard deviation (\(\sigma\)) of 20, we need to calculate the z-scores for both values (45 and 83).
The formula to calculate the z-score is:
\[
z = \frac{(X - \mu)}{\sigma}
\]
### Step 1: Calculate the z-score for 45 apps
For \(X = 45\):
\[
z_{45} = \frac{(45 - 80)}{20} = \frac{-35}{20} = -1.75
\]
### Step 2: Calculate the z-score for 83 apps
For \(X = 83\):
\[
z_{83} = \frac{(83 - 80)}{20} = \frac{3}{20} = 0.15
\]
### Step 3: Look up the z-scores in the standard normal distribution table
Using the standard normal distribution table or a calculator:
- For \(z = -1.75\), the area to the left is approximately \(0.0401\).
- For \(z = 0.15\), the area to the left is approximately \(0.5596\).
### Step 4: Calculate the proportion of phones with apps between 45 and 83
To find the proportion of phones with between 45 and 83 apps, subtract the area to the left of \(z = -1.75\) from the area to the left of \(z = 0.15\):
\[
P(45 < X < 83) = P(Z < 0.15) - P(Z < -1.75)
\]
\[
P(45 < X < 83) = 0.5596 - 0.0401 = 0.5195
\]
### Final Result
The proportion of phones that have between 45 and 83 apps is approximately \(0.5195\).
Thus, rounding to four decimal places:
\[
\boxed{0.5195}
\]
The formula to calculate the z-score is:
\[
z = \frac{(X - \mu)}{\sigma}
\]
### Step 1: Calculate the z-score for 45 apps
For \(X = 45\):
\[
z_{45} = \frac{(45 - 80)}{20} = \frac{-35}{20} = -1.75
\]
### Step 2: Calculate the z-score for 83 apps
For \(X = 83\):
\[
z_{83} = \frac{(83 - 80)}{20} = \frac{3}{20} = 0.15
\]
### Step 3: Look up the z-scores in the standard normal distribution table
Using the standard normal distribution table or a calculator:
- For \(z = -1.75\), the area to the left is approximately \(0.0401\).
- For \(z = 0.15\), the area to the left is approximately \(0.5596\).
### Step 4: Calculate the proportion of phones with apps between 45 and 83
To find the proportion of phones with between 45 and 83 apps, subtract the area to the left of \(z = -1.75\) from the area to the left of \(z = 0.15\):
\[
P(45 < X < 83) = P(Z < 0.15) - P(Z < -1.75)
\]
\[
P(45 < X < 83) = 0.5596 - 0.0401 = 0.5195
\]
### Final Result
The proportion of phones that have between 45 and 83 apps is approximately \(0.5195\).
Thus, rounding to four decimal places:
\[
\boxed{0.5195}
\]
(b) Find the 60th percentile of the number of apps. Round the answer to two decimal places.
GPT-4o mini
To find the 60th percentile of the number of apps, we first need to determine the z-score that corresponds to the 60th percentile in the standard normal distribution.
### Step 1: Find the z-score for the 60th percentile
The z-score for the 60th percentile can be found using a standard normal distribution table or a calculator. The z-score corresponding to 0.60 (or the 60th percentile) is approximately:
\[
z \approx 0.2533
\]
### Step 2: Use the z-score to find the value in the original distribution
We can then use the z-score formula to find \(X\):
\[
z = \frac{(X - \mu)}{\sigma}
\]
Rearranging this formula, we have:
\[
X = z \cdot \sigma + \mu
\]
Substituting in our values:
- \(\mu = 80\) (mean)
- \(\sigma = 20\) (standard deviation)
- \(z \approx 0.2533\)
Calculating \(X\):
\[
X = (0.2533 \cdot 20) + 80
\]
\[
X = 5.066 + 80
\]
\[
X \approx 85.07
\]
### Final Result
Rounding to two decimal places, the 60th percentile of the number of apps on a smartphone is:
\[
\boxed{85.07}
\]
### Step 1: Find the z-score for the 60th percentile
The z-score for the 60th percentile can be found using a standard normal distribution table or a calculator. The z-score corresponding to 0.60 (or the 60th percentile) is approximately:
\[
z \approx 0.2533
\]
### Step 2: Use the z-score to find the value in the original distribution
We can then use the z-score formula to find \(X\):
\[
z = \frac{(X - \mu)}{\sigma}
\]
Rearranging this formula, we have:
\[
X = z \cdot \sigma + \mu
\]
Substituting in our values:
- \(\mu = 80\) (mean)
- \(\sigma = 20\) (standard deviation)
- \(z \approx 0.2533\)
Calculating \(X\):
\[
X = (0.2533 \cdot 20) + 80
\]
\[
X = 5.066 + 80
\]
\[
X \approx 85.07
\]
### Final Result
Rounding to two decimal places, the 60th percentile of the number of apps on a smartphone is:
\[
\boxed{85.07}
\]