statistical procedures are working efficiently. A sample of 800 units has been taken for observations for the study while operations were known to be operating at satisfactory levels. .21 has been set as the sample standard deviation and likewise so has the population standard deviation assumed to be .21 because there are a substantial amount of observations. Random samples of 30 will allow for swift action if said processes are not producing at efficient levels or drop below an average of 12 units‚ the
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We can do a lot of good statistics with the normal curve‚ but virtually none with any other curve. Let us assume that we have recorded the 1000 ages and computed the mean and standard deviation of these ages. Assuming the mean age came out as 40 years and the standard deviation as
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heights of boys at a particular age follows a normal distribution with mean 150.3 cm and standard deviation 5 cm. Find the probability that a boy picked at random from this age group has height (a) less than 153 cm (b) less than 148 cm (c) more than 158cm (d) more than 144 cm (e) between 147 cm and 149.5cm (f) between 150 cm and 158 cm 2. The mean mark on a final examination was 72 and the standard deviation was 9. The top 10% of the students are to receive A’s. What is the minimum mark a student must
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acceptable performance. Also there is the lower specification limit (LSL)‚ which identifies the lowest amount that could give acceptable performance of the process. The USL and LSL are also known as the control limits that are either 3 deviations above the mean or 3 deviations below it (Chase‚ Jacobs‚ and Aquilliano‚ 2006). In developing a control chart for the process for getting ready for work‚ one
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different temperatures. After this‚ equipment called colorimeter will be used to take a reading for absorbency. The readings for the absorbency will be repeated three times in order to find the mean of each sample and to also find out the standard deviation of each set of data which will be recorded in a table format. The temperature rising will cause damage to the cell membrane due to the temperature rising above the membranes temperature. The mean absorbency will be calculated from five different
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CHAPTER 22 estimating risk and return on assets 1. WHAT IS RISK? Risk is the variability of an asset’s future returns. When only one return is possible‚ there is no risk. When more than one return is possible‚ the asset is risky. The greater the variability‚ the greater the risk. 2. RISK – RETURN RELATIONSHIP Investment risk is related to the probability of actually earning less than the expected return – the greater the chance of low or negative returns‚ the riskier the investment
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1.if selecting samples of size n = 10 from a population with a known mean and standard deviation what requirement‚ if any must be satisfied in order to assume that the distribution of the sample means is a normal distribution The population must have a normal distribution. 2. find the area of the shaded region. The graph depicts that standard normal distribution with mean 0 and standard deviation 1. M: 0 δ: 1 Z: 1.13= .8708 2ND DIST. #2 LOWER: -999999 UPPER: 1.13 U: 0 δ: 1 =.8707618393 3. Shaded
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standard deviation and the Z- score for stress among males and females employees working in industries. Variables | N | Mean | Standard Deviation | Z- score | Males | 100 | 139.33 | 18.276 | 2.525 | Females | 100 | 146.08 | 19.514 | | Table 4.1.1. Illustrates the mean‚ standard deviation and the Z-score for stress of 100 males and 100 females (N=200). As shown in the table the mean score for stress among males and females is139.33 and 146.08 respectively‚ with a standard deviation of 18.276
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intervals verses point estimates? The sample mean is a point estimate (single number estimate) of the population mean – Due to sampling error‚ we know this is off. Instead‚ we construct an interval estimate‚ which takes into account the standard deviation‚ and sample size. – Usually stated as (point estimate) ± (margin of error) • What is meant by a 95% confidence interval? That we are 95% confident that our calculated confidence interval actually contains the true mean. • What is the logic
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length of labor in this study descriptive‚ this being represented by mean and standard deviation in the data. Yes‚these are appropriate as they both can be calculated (varible n=30 and mean=14.63) at the interval level of measurements. 3) Nominal level data could be used to describe the length of labor. This would have done by using frequencies‚ percentages‚ and mode instead of using mean‚ range‚ and standard deviation 4) No‚ the distributions of scores were not similar for the two groups. Experimental
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