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KengaRu [80]
3 years ago
12

HELLPPPP MEEEEE ASAPPPPPPP PLEASEEEEEEEEEEEEWEEEEEEEE​

Mathematics
2 answers:
emmasim [6.3K]3 years ago
6 0

Answer:SORRY CANT HELP U THERE

Step-by-step explanation:

HACTEHA [7]3 years ago
6 0

Answer:

that's rough buddy

Step-by-step explanation:

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2 years ago
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Please!! 10 points but adding 5 more!! so 15 points!!
ololo11 [35]
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3 years ago
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[Q71 Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters g = 71 inch and 02 = 6
viktelen [127]

Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

              (b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

Step-by-step explanation:

Given that,

                  Height (in inches) of a 25 year old man is a normal random variable with mean g=71 and variance o^{2} =6.25.

To find:  (a) What percentage of 25 year old men are 6 feet, 2 inches tall

               (b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.

Now,

(a) To calculate the percentage of men, we have to calculate the probability

P[Height of a 25 year old man is over 6 feet 2 inches]= P[X>74in]

                           P[X>74] = P[\frac{X-g}{o} > \frac{74-71}{2.5}]

                                         = P[Z > 1.2]

                                         = 1 - P[Z ≤ 1.2]

                                         = 1 - Ф (1.2)

                                         = 1 - 0.8849

                                         = 0.1151

Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.

(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X, 6ft 5in - X, 6ft]

     P[X > 6ft 5in I X > 6ft] = P[X > 77 I X > 72]

                                          = \frac{P[X > 77]}{P[ X > 72]}

                                          = \frac{P[\frac{X - g}{o}>\frac{77-71}{2.5}]  }{P[\frac{X-g}{o} >\frac{72-71}{2.5}] }

                                          = \frac{P[Z >2.4]}{P[Z>0.4]}

                                          =  \frac{1-P[Z\leq2.4] }{1-P[Z\leq0.4] }

                                          = \frac{1-0.9918}{1-0.6554}

                                          = \frac{0.0082}{0.3446}

                                          = 0.024

Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.

4 0
3 years ago
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