Use the following information for questions 34-36: Deanna is the principal at a Midwestern middle school and wants to know the average IQ of all female, seventh grade students. She does not know anything about what the population distribution looks like. She took a simple random sample of 31 seventh-grade girls in her school and found the average IQ score in her sample was 105.8 and the standard deviation was 15. Based on the interval you calculated in question 34, does it seem plausible that the true average IQ score for all seventh-grade female students at this school is 113

Since, 113 is on the confidence interval obtained (97.502, 114.098), so, we can suggest that is plausible that the true average IQ score for all seventh-grade female students at this school is 113.

Step-by-step explanation:

The question isn't complete, the missing part asked us to obtain a 99.5% confidence interval for the true average IQ score

Finding the confidence interval using the sample data provided, we can answer the question of plausibility.

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 105.8

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 31 - 1 = 30

Significance level for 99.5% confidence interval

(100% - 99.5%)/2 = 0.25% = 0.0025

t (0.0025, 30) = 3.03 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 15

n = sample size = 30

σₓ = (15/√30) = 2.739

99.5% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 105.8 ± (3.03 × 2.739)

CI = 105.8 ± 8.298

99.5% CI = (97.502, 111.387)

99.5% Confidence interval = (97.502, 114.098)

Since, 113 is on the confidence interval obtained, so, we can suggest that is plausible that the true average IQ score for all seventh-grade female students at this school is 113.

Hope this Helps!!!

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb

94

step-by-step explanation:

average the numbers and use a variable to figure out the minimum to get a 90.

(79+94+91+92+x)/5=90

x = 94