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3 years ago

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Question Help Scores of an IQ test have a bell-shaped distribution with a mean of 100100 and a standard deviation of 1010. Use

the empirical rule to determine the following. (a) What percentage of people has an IQ score between 8080 and 120120? (b) What percentage of people has an IQ score less than 9090 or greater than 110110? (c) What percentage of people has an IQ score greater than 120120?

1 answer:

3241004551 [841]3 years ago

6 0

Answer:

a) 95% percentage of people has an IQ score between 80 and 120.

b) 68% percentage of people has an IQ score between 90 and 110.

c) 2.5% percentage of people has an IQ score greater than 120.

Step-by-step explanation:

The empirical formula states that:

The empirical rule is known as the three-sigma rule or 68-95-99.7 rule.
It is a rule which states that for a normal distribution, almost all data falls within three standard deviations of the mean.
68% of data falls within the first standard deviation that is $\mu \pm \sigma$

95% of data falls within the second standard deviation that is $\mu \pm 2\sigma$

99.7% of data falls within the third standard deviation that is $\mu \pm 3\sigma$

a) percentage of people has an IQ score between 80 and 120

(80, 120) can be written as:

$(80, 120) = (100-2(10), 100 + 2(10))\\= (\mu \pm 2\sigma)$

Thus, it is the interval within two standard deviation from the mean.

Thus, by empirical formula, 95% percentage of people has an IQ score between 80 and 120.

b) percentage of people has an IQ score between 90 and 110

(90, 110) can be written as:

$(90, 110) = (100-1(10), 100 + 1(10))\\= (\mu \pm 1\sigma)$

Thus, it is the interval within one standard deviation from the mean.

Thus, by empirical formula, 68% percentage of people has an IQ score between 90 and 110.

c) P(score greater than 120)

= 1 - percentage of score less than 80 - percentage of score between 80 and 120

$= 1 - (95\% + 2.5\%) = 2.5\%$

From empirical formula,

2.5% percentage of people has an IQ score greater than 120.

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Answer:

Step-by-step explanation:

Given that:

A study is conducted and the study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands.

a)

Thus; we formulate the null and the alternative hypotheses as follows:

The proportion is ; $\frac{64}{100}$ = 0.64

Null hypothesis: ${H_0}:p = 0.64$

The Null hypothesis states that there is no evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.

Alternative hypothesis: ${H_a}:p \ne 0.64H$

The Alternative hypothesis states that there is evidence that “the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%”.

b)

The proportion of the population = 0.64

The sample size n = 100

The number of shoppers = 52 stating that the supermarket brand was as good as the national brand

The sample proportion $\hat p = \frac{52}{100}$

$\hat p = 0.52$

The z - value is calculated by the formula:

$z = \dfrac{\hat p-p}{\frac{\sqrt{ p(1-p)}}{100 }}$

$z = \dfrac{0.52-0.64}{\frac{\sqrt{ 0.64(1-0.64)}}{100 }}$

$z = \dfrac{-0.12}{0.048 }}$

z = -2.50

Since z = -2.50

the p-value = 2P(Z ≤ -2.50)

p-value = 2 × 0.0062

The p-value = 0.0124

c)

At significance level, ∝ = 0.05 ; The p-value = 0.0124

According to the rejection rule, if p-value is less than 0.05 then we will reject null hypothesis at ∝ = 0.05

Hence, the p-value =0.0124 < ∝ (=0.05)

According to the reject rule; reject null hypothesis.

Conclusion: There is no evidence at all that the percentage of supermarket shoppers believes that the supermarket ketchup is good as the national brand ketchup differ form 64%.

d) we know that the significance level of ∝ = 0.05

The value ${z__{0.05}}$ is obtained as:

$P\left( {\left| Z \right| \le z} \right) = 0.05$

Now; to determine Z ; we locate the probability value of 0.05 form the table of standard normal distribution. Then we proceed to the left until the first column is reached and the value is 1.90. Also , we move upward until we reach the top and the value is 0.06. Now; the intersection of the row and column results the area to the left of z

This implies that : $P(Z \leq -1.96)=0.05$

The critical value for left tail is -1.96 and the critical value for right tail is 1.96.

Conclusion:

The critical value is -1.96 and the value of test statistic is - 2.50. Here, we can see that the value of test statistic is lesser than the critical value. Hence, we can be concluded that there is evidence that reject the null hypothesis.

Therefore, Yes, the national brand ketchup manufacturer is pleased with the conclusion.

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Answer:

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Formula

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Substitution

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