Web Calculator Exercise 1

The area under the curve to the left of the unknown quantity must be 0.7 (70%). So, we must first find the z value that cuts off an area of 0.7 in the left tail of standard normal distribution. Using the cumulative probability table, we see that z=0.53.

2. In order to determine the average amount spent in November on Amazon.com a random sample of 144 Amazon accounts were selected. The sample mean amount spent in November was $250 with a standard deviation of $25. Assuming that the population standard deviation is unknown, what is a 95% confidence interval for the population mean amount spent on Amazon.com in November?

(b) The value of Z with an area of 5% in the right tail, but not the sample mean.

1. By hand, compute the mean, median, and mode for the following set of 40 reading scores:

If John gets an 90 on a physics test where the mean is 85 and the standard deviation is 3, where does he stand in relation to his classmates? (he is in the top 5%, he is in the top 10%, he is in the bottom 5%, or bottom 1%)

Slide 15: Here’s the same curve demonstrating the z-score values for 2.5% at each tail.

d. What is the dealer’s final price? e. What is the dealer’s total savings? III. Equation a. Rebate = Original Price – Dealer Price b. %

σA = 0.3 × (0.07)2 + 0.4 × (0.06)2 + 0.3 × (0.08)2 − (0.021)2 = 0.004389,

1. Why is a z score a standard score? Why can standard scores be used to compare scores from different distributions? It is a scores relationship to the mean indicating whether it is above or below the mean. It does this by converting scores to z score. Yes – keep going – just a bit more is needed.2 out of 3 pts

The variable age is the independent variable and is a ratio level of measurement (Loiselle et al., 2011). The measure of central tendency to describe age are in table 1.2 are the mean of 57.62 which is the average age, the median which is the middle score within the distribution when all scores are organized of 58.5 and the mode of 58 which is the most frequently occurring age (Loiselle et al., 2011). The measures of variability are the range of 69 with a minimum age of 22 and a maximum age of 91, standard deviation which is the average deviation from the sample mean which is a value of 16.26, and the sample variance which is the standard deviation square and the value is 263.46 (Salkind, 2013). The distribution for this sample is described as a negative skew and the value obtained from table 1.2 is -0.22511(Salkind, 2013). A negative skew occurs when the median and the mode value are larger than the mean, within this sample the median is 58.5 the mode is 58 which is greater than the mean of 57.62, the tail would be pointed toward the left (Salkind, 2013). The kurtosis value is -0.65102 and this describes how peak or flat the curve is from the normal distribution curve which is described as mesokurtic (Salkind, 2013). The kurtosis has a large negative value which is representative of a flatter curve also know as playkurtic (Salkind, 2013).

3. Use the tables in the text to determine the critical value used to conduct the test. If there are two critical values, state only the upper value.

In the plot above we are able to come to various conclusions. The first points we have to analyze are our quartiles:

Strictly speaking, the lower the score, the higher the odds are that a company is headed for bankruptcy. A Z-score of lower than 1.8 in particular, indicates that the company is heading for bankruptcy. Companies with scores above 3 are unlikely to enter bankruptcy. Scores in between 1.8 and 3 lie in a gray area.

Of the 22 students surveyed on miles with range of 68 the mean is 21.34, Median is 15, Mode is 2 and standard deviation is 3.87.

Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table)