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)?&#10;A) 5.8&#10;B) 5.9&#10;C) 6.0&#10;D) 6.1

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?&#10;A) 0.4772&#10;B) 0.5&#10;C) 0.9772&#10;D) 0.0228

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).

A) AED 697.3
B) AED 704.1
C) AED 723.6
D) AED 734.9

AED 704.1

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)&#10;A) 24.5 &#177; 0.8&#10;B) 24.5 &#177; 0.9&#10;C) 24.5 &#177; 1.0&#10;D) 24.5 &#177; 1.1

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&#10;A) 2.0&#10;B) 2.2&#10;C) 2.5&#10;D) 2.8

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?&#10;A) The normal distribution is symmetrically shaped&#10;B) The normal distribution is centred around the mean&#10;C) The normal distribution shows a clustering of values around the mean&#10;D) The normal distribution is not suitable for use with continuous data

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?&#10;A) Mean&#10;B) Z-score&#10;C) Standard Deviation&#10;D) Variance

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?

A) No, -1.96 < z < 1.96
B) Yes, they have gone down, z < -1.96
C) Yes, they have gone up z > 1.96
D) There is no evidence of a significant difference

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)?

A) 0.12 ± 0.06
B) 0.12 ± 0.09
C) 0.12 ± 0.12
D) 0.12 ± 0.15

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)?

A) 5
B) 6
C) 7
D) 8

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.

Deck 34: Normal Distribution