a. For a smartphone app distribution with a mean of 85 and a standard deviation of 30, approximately 35.55% of phones have between 83 and 96 apps.
b. The 65th percentile corresponds to around 96.559 apps.
To solve these problems, we can use the standard normal distribution (Z) and the Z-table. The Z-score is calculated using the formula:
\[ Z = \frac{{X - \mu}}{{\sigma}} \]
where:
- \( X \) is the value,
- \( \mu \) is the mean,
- \( \sigma \) is the standard deviation.
For Part 1:
(a) To find the proportion of phones with apps between 83 and 96, we need to find the Z-scores for these values and then find the area between those Z-scores in the Z-table.
\[ Z_{83} = \frac{{83 - 85}}{{30}} \]
\[ Z_{96} = \frac{{96 - 85}}{{30}} \]
Now, use the Z-table to find the area between \( Z_{83} \) and \( Z_{96} \).
For Part 2:
(b) To find the 65th percentile, we need to find the Z-score corresponding to the 65th percentile and then use it to find the corresponding value.
\[ Z_{65\%} \]
Now, use the Z-table to find the value corresponding to \( Z_{65\%} \).
Please note that I'll provide the calculated answers in the next response due to message length limitations.
For Part 1:
(a) Calculate the Z-scores:
\[ Z_{83} = \frac{{83 - 85}}{{30}} = -0.0667 \]
\[ Z_{96} = \frac{{96 - 85}}{{30}} = 0.3667 \]
Now, use the Z-table to find the area between \( Z_{83} \) and \( Z_{96} \):
\[ P(83 < X < 96) = P(-0.0667 < Z < 0.3667) \]
Consulting the Z-table, we find the corresponding values:
\[ P(-0.0667 < Z < 0.3667) \approx 0.3555 \]
So, approximately 35.55% of phones have between 83 and 96 apps.
For Part 2:
(b) To find the Z-score corresponding to the 65th percentile:
\[ P(Z < Z_{65\%}) = 0.65 \]
Looking up this value in the Z-table gives \( Z_{65\%} \approx 0.3853 \).
Now, use the Z-score formula to find the corresponding value:
\[ X_{65\%} = \mu + Z_{65\%} \times \sigma \]
\[ X_{65\%} = 85 + 0.3853 \times 30 \]
\[ X_{65\%} \approx 96.559 \]
So, the 65th percentile of the number of apps is approximately 96.559.