Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. Find P25,which is the IQ score separating the bottom 25% from the top 75%. Round your answer to nearest tenth. ---### Understanding IQ Scores and the 25th PercentileAssume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. **Problem Statement:**Find $ P_{25} $, which is the IQ score separating the bottom 25% from the top 75%. Round your answer to the nearest tenth.**Options:**- A. 91.2- B. 91.8- C. 90.5- D. 89.9Click below to view the respective pages of the table necessary for this calculation:- [Click to view page 1 of the table](#)- [Click to view page 2 of the table](#)**Countdown Timer:**- Time Remaining: 03:39:16**Navigation:**- [Next](#)**Additional Resources:**- [1102Research_p_...doc](#)- [thumbnail_IMG_...jpg](#)- [Tables-Z-T-P-Va....pdf](#)---Explanation of the Problem:In the context of this problem, we aim to identify the IQ score (denoted as $ P_{25} $) that splits the bottom 25% of the population from the top 75%. This is a typical application of the percentiles in a normal distribution. To find this, we look up a Z-table or use statistical software to determine the specific IQ score corresponding to the 25th percentile of a normal distribution.Given:- Mean (μ) = 100- Standard Deviation (σ) = 15This enables us to convert between Z-scores and IQ scores using the formula:\[ IQ = μ + Z \times σ \]Remember to verify your answer by checking the provided choices accurately.---

The provided information is µ=100 σ=15 P(X<x)=0.25P(X-µσ<x-10015)=0.25P(z<-0.6745)=0.25…

Answered: Assume that adults have IQ scores that…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.4: Distributions Of Data

Problem 22PFA

See similar textbooks

Related questions

Question

Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. Find P25,

which is the IQ score separating the bottom 25% from the top 75%. Round your answer to nearest tenth.

---

### Understanding IQ Scores and the 25th Percentile

Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15.

**Problem Statement:**
Find $ P_{25} $, which is the IQ score separating the bottom 25% from the top 75%. Round your answer to the nearest tenth.

**Options:**
- A. 91.2
- B. 91.8
- C. 90.5
- D. 89.9

Click below to view the respective pages of the table necessary for this calculation:

- [Click to view page 1 of the table](#)
- [Click to view page 2 of the table](#)

**Countdown Timer:**
- Time Remaining: 03:39:16

**Navigation:**
- [Next](#)

**Additional Resources:**
- [1102Research_p_...doc](#)
- [thumbnail_IMG_...jpg](#)
- [Tables-Z-T-P-Va....pdf](#)

---

Explanation of the Problem:
In the context of this problem, we aim to identify the IQ score (denoted as $ P_{25} $) that splits the bottom 25% of the population from the top 75%. This is a typical application of the percentiles in a normal distribution. To find this, we look up a Z-table or use statistical software to determine the specific IQ score corresponding to the 25th percentile of a normal distribution.

Given:
- Mean (μ) = 100
- Standard Deviation (σ) = 15

This enables us to convert between Z-scores and IQ scores using the formula:
\[ IQ = μ + Z \times σ \]

Remember to verify your answer by checking the provided choices accurately.

---

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution