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your revlew only The lengths of pregnancies in a small rural village are normally distributed with...
The lengths of pregnancies in a small rural village are normally distributed with a mean of...
The lengths of pregnancies in a small rural village are normally distributed with a mean of 260 days and a standard deviation of 15 days. What percentage of pregnancies last fewer than 295 days? PIX < 295 days) = % Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Submit Question
The lengths of pregnancies in a small rural village are normally distributed with a mean of...
The lengths of pregnancies in a small rural village are normally distributed with a mean of 269 days and a standard deviation of 13 days. A distribution of values is normal with a mean of 269 and a standard deviation of 13. What percentage of pregnancies last beyond 276 days? P(X > 276 days) = ____% Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores...
The lengths of pregnancies in a small rural village are normally distributed with a mean of...
The lengths of pregnancies in a small rural village are normally distributed with a mean of 264 days and a standard deviation of 12 days. In what range would you expect to find the middle 50% of most pregnancies? Between and . If you were to draw samples of size 38 from this population, in what range would you expect to find the middle 50% of most averages for the lengths of pregnancies in the sample? Between and . Enter...
The lengths of pregnancies in a small rural village are normally distributed with a mean of...
The lengths of pregnancies in a small rural village are normally distributed with a mean of 264 days and a standard deviation of 13 days. A distribution of values is normal with a mean of 264 and a standard deviation of 13. What percentage of pregnancies last fewer than 294 days? P(X < 294 days) = % ?
The lengths of pregnancies in a small rural village are normally distributed with a mean of...
The lengths of pregnancies in a small rural village are normally distributed with a mean of 265 days and a standard deviation of 16 days. In what range would you expect to find the middle 68% of most pregnancies? Between and . If you were to draw samples of size 48 from this population, in what range would you expect to find the middle 68% of most averages for the lengths of pregnancies in the sample? Between and . Enter your answers...
The lengths of pregnancies in a small rural village are normally distributed with a mean of...
The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 14 days. In what range would you expect to find the middle 95% of most pregnancies? Between 233 xan x and 284 x! If you were to draw samples of size 40 from this population, in what range would you expect to find the middle 95% of most averages for the lengths of pregnancies in the sample?...
The lengths of pregnancies in a small rural village are normally distributed with a mean of...
The lengths of pregnancies in a small rural village are normally distributed with a mean of 264 days and a standard deviation of 15 days. In what range would you expect to find the middle 68% of most pregnancies? Between and . If you were to draw samples of size 47 from this population, in what range would you expect to find the middle 68% of most averages for the lengths of pregnancies in the sample? Between and .
1) The lengths of pregnancies in a small rural village are normally distributed with a mean...
1) The lengths of pregnancies in a small rural village are normally distributed with a mean of 263 days and a standard deviation of 15 days. A distribution of values is normal with a mean of 263 and a standard deviation of 15. What percentage of pregnancies last fewer than 295 days? P(X < 295 days) = % 2) The physical fitness of an athlete is often measured by how much oxygen the athlete takes in (which is recorded in...
A company produces steel rods. The lengths of the steel rods are normally distributed with a...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 220.3-cm and a standard deviation of 2.2-cm. Find P22, which is the length separating the shortest 33% rods from the longest 67%. P33 = 1.94 x-cm Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
company produces steel rods. The lengths of the steel rods are normally distributed with a mean...
company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 203.5-cm and a standard deviation of 2.4-cm. For shipment, 7 steel rods are bundled together. Find P94, which is the average length separating the smallest 94% bundles from the largest 6% bundles. P94 = -cm Enter your answer as a number accurate to 2 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
The lengths of pregnancies in a small rural village are normally distributed with a mean of 260 days and a standard deviation of 15 days. What percentage of pregnancies last fewer than 295 days? PIX < 295 days) = % Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Submit Question
The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 14 days. In what range would you expect to find the middle 95% of most pregnancies? Between 233 xan x and 284 x! If you were to draw samples of size 40 from this population, in what range would you expect to find the middle 95% of most averages for the lengths of pregnancies in the sample?...
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 220.3-cm and a standard deviation of 2.2-cm. Find P22, which is the length separating the shortest 33% rods from the longest 67%. P33 = 1.94 x-cm Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
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