Question
A:) For a standard normal distribution (use the normal table), find: P(z>0.65)P(z>0.65), the percent of data values above z=0.65z=0.65. The normal distribution table gives an
A:) For a standard normal distribution (use the normal table), find:
P(z>0.65)P(z>0.65), the percent of data values above z=0.65z=0.65.
The normal distribution table gives an area in decimal form for P(z>0.65)=P(z>0.65)= .
Converted to percent (%), the percent of data values above z=0.65z=0.65 is P(z>0.65)=P(z>0.65)= %.
B:)For a standard normal distribution (using the normal distribution table), find P(0.23 The normal distribution table gives the decimal area between z=0.23z=0.23 and z=2.19z=2.19 as P(0.23 Converted to percent (%), the percent of data values between z=0.23z=0.23 and z=2.19z=2.19 is P(0.23 C:)A particular fruit's weights are normally distributed, with a mean of 767 grams and a standard deviation of 31 grams. If you pick one fruit at random, what is the probability that it will weigh between 799 grams and 805 grams? D:)The heights of adult men in America are normally distributed, with a mean of 69.4 inches and a standard deviation of 2.65 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.3 inches and a standard deviation of 2.52 inches. a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)? z = b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent. % c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)? z = d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent. % e) Who is relatively taller: a 6'3" American man or a 5'11" American woman? Defend your choice in a meaningful sentence.
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