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Answer:
1. 15.87%
2. 6 pounds and 8.8 pounds.
3. 2.28%
4. 50% of newborn babies weigh more than 7.4 pounds.
5. 84%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 7.4 pounds
Standard Deviation, σ = 0.7 pounds
We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
1.Percent of newborn babies weigh more than 8.1 pounds
P(x > 8.1)
[tex]P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)[/tex]
[tex]= 1 - P(z \leq 1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%[/tex]
15.87% of newborn babies weigh more than 8.1 pounds.
2.The middle 95% of newborn babies weight
Empirical Formula:
- Almost all the data lies within three standard deviation from the mean for a normally distributed data.
- About 68% of data lies within one standard deviation from the mean.
- About 95% of data lies within two standard deviations of the mean.
- About 99.7% of data lies within three standard deviation of the mean.
Thus, from empirical formula 95% of newborn babies will lie between
[tex]\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8[/tex]
95% of newborn babies will lie between 6 pounds and 8.8 pounds.
3. Percent of newborn babies weigh less than 6 pounds
P(x < 6)
[tex]P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 6) =0.0228 = 2.28\%[/tex]
2.28% of newborn babies weigh less than 6 pounds.
4. 50% of newborn babies weigh more than pounds.
The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.
Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.
5. Percent of newborn babies weigh between 6.7 and 9.5 pounds
[tex]P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%[/tex]
84% of newborn babies weigh between 6.7 and 9.5 pounds.
Using the normal distribution, it is found that:
1. 15.87% of newborn babies weigh more than 8.1 pounds.
2. The middle 95% of newborn babies weigh between 6.03 and 8.77 pounds.
3. 2.28% of newborn babies weigh less than 6 pounds.
4. Approximately 50% of newborn babies weigh more than 7.4 pounds.
5. 84% of newborn babies weigh between 6.7 and 9.5 pounds.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of 7.4 pounds, hence [tex]\mu = 7.4[/tex]
- The standard deviation is of 0.7 pounds, hence [tex]\sigma = 0.7[/tex].
Item 1:
The proportion is 1 subtracted by the p-value of Z when X = 8.1, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{8.1 - 7.4}{0.7}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.8413.
1 - 0.8413 = 0.1587.
0.1587 = 15.87% of newborn babies weigh more than 8.1 pounds.
Item 2:
Between the (100 - 95)/2 = 2.5th percentile and the (100 + 95)/2 = 97.5th percentile.
2.5th percentile:
X when Z has a p-value of 0.025, so X when Z = -1.96.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.96 = \frac{X - 7.4}{0.7}[/tex]
[tex]X - 7.4 = -1.96(0.7)[/tex]
[tex]X = 6.03[/tex]
97.5th percentile:
X when Z has a p-value of 0.975, so X when Z = 1.96.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.96 = \frac{X - 7.4}{0.7}[/tex]
[tex]X - 7.4 = 1.96(0.7)[/tex]
[tex]X = 8.77[/tex]
The middle 95% of newborn babies weigh between 6.03 and 8.77 pounds.
Item 3:
The proportion is the p-value of Z when X = 6, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{6 - 7.4}{0.7}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a p-value of 0.0228.
0.0228 = 2.28% of newborn babies weigh less than 6 pounds.
Item 4:
Due to the symmetry of the normal distribution, 50% of newborn babies weigh more than the mean, which is of 7.4 pounds.
Item 5:
The proportion is the p-value of Z when X = 9.5 subtracted by the p-value of Z when X = 6.7, hence:
X = 9.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9.5 - 7.4}{0.7}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a p-value of 0.9987.
X = 6.7
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{6.7 - 7.4}{0.7}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a p-value of 0.1587.
0.9987 - 0.1587 = 0.84
0.84 = 84% of newborn babies weigh between 6.7 and 9.5 pounds.
A similar problem is given at https://brainly.com/question/24663213
