The distribution of heights of 19 -year-old women is approximately normal, with a mean of 170 cm and a standard deviation of 5 cm . a What percentage of these women have heights: i between 155 and 185 cm ? ii greater than 180 cm ? iii between 160 and 175 cm ? b In a sample of 5000 of these women, approximately how many have heights greater than 175 cm ?

A person's blood glucose level and diabetes are dosely related. Let \( x \) be a random variable measured in milligrams of glucose per deciliter ( \( 1 / 10 \) of a liter) of blood. Suppose that after a 12 -hoi fast, the random variable \( x \) will have a distribution that is approximately normal with mean \( \mu=87 \) and standard deviation \( \sigma=23 \). Note: After 50 years of age, both the mean and standard deviation tend to increase. For an adult (under 50 ) after a 12-hour fast, find the following probabilities. (Round your answers to four decimal places.) (a) \( x \) is more than 60 (b) \( x \) is less than 110 (c) \( x \) is between 60 and 110 (d) \( x \) is greater than 125 (borderline diabetes starts at 125 )

The test scores of 40 students are listed below. Find \( \mathrm{P}_{85} \). \( \begin{array}{l}30,35,43,44,47,48,54,55,56,57,59,62,63,65,66,68,69,69,71,72,72,73, \\ 74,76,77,77,78,79,80,81,81,82,83,85,89,92,93,94,97,98\end{array} \)

The regression equation is \( \bar{y}=4.2 \) decimal places if necessary.) For \( x=5.8, \hat{y}=74.66 \) (Round to two decimal places if necessary.) For \( x=11.5, \hat{y}=98.6 \) The prenatal care percent for 11.5 interpolation, while the one for 5.8

Question 8 The heights of 10 -year-olds, regardless of gender, closely follow a Normal distribution with a mean of 56 inches and a standard deviation of 5 inches. a. What is the probability that a randomly chosen 10 -year-old is shorter than 48 inches? (Round to 4 decimal places.) b. If the tallest \( 10 \% \) of the class is considered "very tall", what is the height cutoff for "very tall"? (Round to 2 decimal places.) [SAB2]

How many college students are needed to estimate the mean grade point average of all current college students in the United Sates within 20\% percentage points with a \( 99 \% \) confidence. An earlier study gave an estimate of 0.79 for the population standard deviation. \( \mathrm{n}= \)