Question 1

If the normal distribution is to be used as an approximation to the binomial with n = 200 and p = 0.78 what value should be used to approximate the standard deviation (1 decimal place)?

Accepted Answer

The standard deviation ($\sigma$) of a binomial distribution can be calculated using the formula $\sigma = \sqrt{n \cdot p \cdot (1-p)}$, where $n$ is the number of trials and $p$ is the probability of success. Substituting $n = 200$ and $p = 0.78$ gives $\sigma = \sqrt{200 \cdot 0.78 \cdot (1-0.78)} = \sqrt{200 \cdot 0.78 \cdot 0.22} \approx 5.9$.

Question 2

If the probability that a Z-score lies between 0 and 2 is 0.452 then what is the probability that Z < 2?

Accepted Answer

The probability that Z < 2 is the sum of the probability that Z lies between 0 and 2 (0.452) and the probability that Z < 0 (0.5, because the normal distribution is symmetric around 0). Therefore, 0.452 + 0.5 = 0.952.

Question 3

Analysis of a sample of 50 of the second quarter sales of a Dubai fashion store shows a difference ( - ) of AED 236, z = +2.37 indicating that the difference is significant. Calculate the standard deviation of the sample (1 decimal place).

Accepted Answer

[note, it is assumed that the book will have a table of Z-score probabilities included - if not, these will need to be added for some questions]

Question 4

Assuming a confidence interval of 95% (z = 1.96), calculate the width of the confidence interval for n = 65, = 24.5, s = 4.5 (1 decimal place)

Accepted Answer

The answer of Assuming a confidence interval of 95% (z...

Question 5

Calculate the Z-score for a value of 34 when mean = 24 and standard deviation = 4

Accepted Answer

The answer of Calculate the Z-score for a value of...

Question 6

Which of the following statements about the normal distribution are NOT true?

Accepted Answer

The normal distribution is indeed suitable for use with continuous data, making statement D incorrect. The other statements (A, B, C) correctly describe characteristics of the normal distribution: it is symmetrically shaped, centered around the mean, and shows a clustering of values around the mean.

Question 7

What name is given to a value that has been standardized?

Accepted Answer

A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, standardized so that it reflects the number of standard deviations away from the mean.

Question 8

Last year average wages in an Indian firm were INR 109,658, a sample of 30 employees in the same firm this year gives an average wage of INR 110,986 (standard deviation INR 1,234). Is this evidence of a significant difference in wages?

Accepted Answer

The z-score is calculated as (110,986 - 109,658) / (1,234/√30) = 5.99, which is greater than 1.96, indicating a significant increase in wages.

Question 9

A sample of components made by a factory in China shows that 6 out of a sample of 50 were found to be faulty. What is the width of a 95% confidence interval (z = 1.96) for the defect rate (2 decimal places)?

Accepted Answer

The defect rate (p) is 6/50 = 0.12. The standard error (SE) for the proportion is calculated using the formula SE = sqrt[(p(1-p))/n], where n is the sample size. SE = sqrt[(0.12*0.88)/50] = 0.046. The margin of error (ME) at 95% confidence level is z*SE, where z is 1.96 for 95% confidence. ME = 1.96*0.046 = 0.09. Therefore, the width of the confidence interval is 0.12 ± 0.09.

Question 10

The diameter of holes drilled into flat-pack furniture is normally distributed with a mean of 5mm and a standard deviation of 0.15mm. If a single item of furniture has 50 holes in it, how many have a diameter of more than 5.2mm (to the nearest whole number)?

Accepted Answer

To find the number of holes with a diameter of more than 5.2mm, we first calculate the z-score for 5.2mm using the formula $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value in question (5.2mm), $\mu$ is the mean (5mm), and $\sigma$ is the standard deviation (0.15mm). This gives us $z = \frac{(5.2 - 5)}{0.15} = \frac{0.2}{0.15} = 1.33$. Looking up 1.33 on the standard normal distribution table gives us an area of approximately 0.9082 to the left of 5.2mm, meaning there's about 9.18% (1 - 0.9082) of the distribution to the right. For 50 holes, 9.18% of 50 is approximately 4.59, which rounds to 5 holes when considering the nearest whole number.

Quiz 34: Normal Distribution

If the normal distribution is to be used as an approximation to the binomial with n = 200 and p = 0.78 what value should be used to approximate the standard deviation (1 decimal place) ?

If the probability that a Z-score lies between 0 and 2 is 0.452 then what is the probability that Z < 2?

Analysis of a sample of 50 of the second quarter sales of a Dubai fashion store shows a difference ( - ) of AED 236, z = +2.37 indicating that the difference is significant. Calculate the standard deviation of the sample (1 decimal place) .

Assuming a confidence interval of 95% (z = 1.96) , calculate the width of the confidence interval for n = 65, = 24.5, s = 4.5 (1 decimal place)

Calculate the Z-score for a value of 34 when mean = 24 and standard deviation = 4

Which of the following statements about the normal distribution are NOT true?

What name is given to a value that has been standardized?

Last year average wages in an Indian firm were INR 109,658, a sample of 30 employees in the same firm this year gives an average wage of INR 110,986 (standard deviation INR 1,234) . Is this evidence of a significant difference in wages?

A sample of components made by a factory in China shows that 6 out of a sample of 50 were found to be faulty. What is the width of a 95% confidence interval (z = 1.96) for the defect rate (2 decimal places) ?

The diameter of holes drilled into flat-pack furniture is normally distributed with a mean of 5mm and a standard deviation of 0.15mm. If a single item of furniture has 50 holes in it, how many have a diameter of more than 5.2mm (to the nearest whole number) ?