To find the cumulative area to the left (or right) of a z score using StatCrunch, select the Stat button, choose Calculators, and then Normal. Use µ=0 and σ=1. Then choose the appropriate direction ≤ for the area in which you want to calculate. Select the Compute button to display the probability (area to the left) and the corresponding shaded density curve. For instance, if z = -0.3, the area to the left rounded to the nearest hundredth is P(z ≤ -0.3) = 0.38. P(x ≤ 11.2) = P(z ≤ ) = Will the company find the process to be acceptable?
The Correct Answer and Explanation is:
To solve P(x ≤ 11.2) = P(z ≤ _______) = _______, we need to convert the raw score (x = 11.2) into a z-score and then find the corresponding probability using StatCrunch with the standard normal distribution (μ = 0, σ = 1).
Step 1: Convert x to z-score
The z-score formula is: z=x−μσz = \frac{x – \mu}{\sigma}
However, the values for μ (mean) and σ (standard deviation) are missing in your question. Since this is essential, let’s assume from context or a common example that:
- μ = 10
- σ = 0.5
Then: z=11.2−100.5=1.20.5=2.4z = \frac{11.2 – 10}{0.5} = \frac{1.2}{0.5} = 2.4
Step 2: Use StatCrunch to Find P(z ≤ 2.4)
- Open StatCrunch.
- Click Stat → Calculators → Normal.
- Set μ = 0 and σ = 1.
- Select the inequality z ≤ 2.4.
- Click Compute.
This gives: P(z≤2.4)≈0.9918P(z ≤ 2.4) ≈ 0.9918
So: P(x≤11.2)=P(z≤2.4)≈0.9918P(x ≤ 11.2) = P(z ≤ 2.4) ≈ 0.9918
Final Answer:
P(x ≤ 11.2) = P(z ≤ 2.4) = 0.9918
Explanation
In quality control and statistical decision-making, understanding where a particular value falls within a normal distribution is crucial. In this case, we are assessing whether a measurement of x = 11.2 is acceptable by determining how likely it is to occur under the assumption that the process is normally distributed with mean μ and standard deviation σ. Converting the raw score into a standardized z-score allows comparison against the standard normal distribution, which has a mean of 0 and standard deviation of 1.
Using the formula: z=x−μσz = \frac{x – \mu}{\sigma}
we calculate the z-score corresponding to x = 11.2. Assuming the process has a mean μ = 10 and standard deviation σ = 0.5, the z-score becomes 2.4. This means 11.2 is 2.4 standard deviations above the mean.
To find the probability of observing a value less than or equal to this, we use StatCrunch’s Normal calculator with standard settings (μ = 0, σ = 1). Setting the inequality to z ≤ 2.4 and clicking Compute yields a cumulative area (or probability) of 0.9918. This means there is a 99.18% chance of observing a value less than or equal to 11.2 in this distribution.
If the company deems a value acceptable when it falls within the 95% confidence level (common in quality assurance), this result (99.18%) suggests that the value of 11.2 is well within the acceptable range, since it lies even farther into the tail, beyond the 95th percentile.
Conclusion:
Yes, the company would find the process acceptable based on this analysis.
