a) Find the number of nails in a bag of 120 that are less than 4.94 in. long. b) Find the probability that the length of nails are between 4.97 and 5.03 in. long. c) What length of... a) Find the number of nails in a bag of 120 that are less than 4.94 in. long. b) Find the probability that the length of nails are between 4.97 and 5.03 in. long. c) What length of nails is considered to be at the 58th percentile of all nails?
Understand the Problem
The question is asking for the analysis of a normally distributed variable (the length of nails) with given parameters (mean and standard deviation). It involves finding the number of nails less than a specific length, calculating a probability of lengths within a range, and determining a specific percentile. These are standard statistical questions related to normal distribution.
Answer
a) 3 nails, b) 0.6826, c) 5.0063 in.
Answer for screen readers
a) 3 nails
b) Probability ≈ 0.6826
c) Length ≈ 5.0063 in.
Steps to Solve
- Calculate the Z-score for 4.94 in.
To find the number of nails less than 4.94 in., we first need to calculate the Z-score using the formula: $$ Z = \frac{X - \mu}{\sigma} $$ where:
- $X = 4.94$ in.
- $\mu = 5.00$ in. (mean)
- $\sigma = 0.03$ in. (standard deviation)
Plugging in the values: $$ Z = \frac{4.94 - 5.00}{0.03} = \frac{-0.06}{0.03} = -2 $$
- Find the probability for the Z-score -2.
Next, we can look up the Z-score of -2 in the standard normal distribution table or use a calculator. This gives us the probability: $$ P(Z < -2) \approx 0.0228 $$
- Calculate the number of nails less than 4.94 in.
To find the number of nails less than 4.94 in. in a bag of 120: $$ \text{Number of nails} = P(Z < -2) \times 120 $$ So, $$ \text{Number of nails} \approx 0.0228 \times 120 \approx 2.736 \approx 3 $$ (rounding to the nearest whole number)
- Calculate the probabilities for the range 4.97 in. to 5.03 in.
Calculate Z-scores for both lengths:
- For $X = 4.97$: $$ Z = \frac{4.97 - 5.00}{0.03} = \frac{-0.03}{0.03} = -1 $$
- For $X = 5.03$: $$ Z = \frac{5.03 - 5.00}{0.03} = \frac{0.03}{0.03} = 1 $$
- Find probabilities for these Z-scores.
Using the standard normal distribution table:
- $P(Z < -1) \approx 0.1587$
- $P(Z < 1) \approx 0.8413$
So the probability between these two values is: $$ P(-1 < Z < 1) = P(Z < 1) - P(Z < -1) $$ Thus, $$ \text{Probability} \approx 0.8413 - 0.1587 = 0.6826 $$
- Find the length at the 58th percentile.
To find the Z-score for the 58th percentile, we can check the Z-table or use a calculator. The Z-score corresponding to 0.58 is approximately: $$ Z \approx 0.21 $$
Calculate the length corresponding to this Z-score: $$ X = \mu + Z \cdot \sigma $$ Substituting in the values: $$ X = 5.00 + 0.21 \cdot 0.03 \approx 5.00 + 0.0063 \approx 5.0063 \text{ in.} $$
a) 3 nails
b) Probability ≈ 0.6826
c) Length ≈ 5.0063 in.
More Information
The calculations revolve around properties of the normal distribution, utilizing Z-scores to interpret probabilities and percentiles. Statistical software or Z-tables can expedite these calculations.
Tips
- Incorrectly applying the Z-score formula (ensure the proper substitution of mean and standard deviation).
- Miscalculating probabilities; always check the Z-table accurately.
- Rounding too early in calculations leading to inaccuracies.
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