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SCORPION-xisa [38]
3 years ago
9

At the Canada Open Tennis Championship, a statistician keeps track of every serve that a player hits during the tournament. The

statistician reported that the mean serve speed of a particular player was 99 miles per hour (mph) and the standard deviation of the serve speeds was 15 mph.
If nothing is known about the shape of the distribution, give an interval that will contain the speeds of at least eight-ninths of the player's serves.

a) 54 mph to 144 mph
b) 39 mph to 159 mph
c) 144 mph to 189 mph
d) 69 mph to 129 mph
Mathematics
1 answer:
kondaur [170]3 years ago
8 0

Answer:

a) 54 mph to 144 mph

Step-by-step explanation:

We don't know the shape of the distribution, so we use Chebyshev's Theorem to solve this question. It states that:

At least 75% of the measures are within 2 standard deviations of the mean.

At least 89% of the measures are within 3 standard deviations of the mean.

At least eight-ninths of the player's serves.

8/9 is approximately 89%

So

Mean: 99, standard deviation: 15

99 - 3*15 = 54

99 + 3*15 = 144

So the correct answer is:

a) 54 mph to 144 mph

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More Information From the Online Dating Survey A survey conducted in July 2015 asked a random sample of American adults whether
svet-max [94.6K]

Answer:

90% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating is [0.095 , 0.148].

Step-by-step explanation:

We are given that a survey conducted in July 2015 asked a random sample of American adults whether they had ever used online dating.

The survey included 411 adults between the ages of 55 and 64, and 50 of them said that they had used online dating.

Firstly, the pivotal quantity for 90% confidence interval for the population proportion is given by;

                                P.Q. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of adults who said that they had used online dating =  \frac{50}{411}  = 0.122

n = sample of adults between the ages of 55 and 64 = 411

p = population proportion of all US adults ages 55 to 64 to use online dating

<em>Here for constructing 90% confidence interval we have used One-sample z proportion statistics.</em>

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                  of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.645) = 0.90

P( -1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

P( \hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

<u>90% confidence interval for p</u> = [\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }, \hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.122-1.645 \times {\sqrt{\frac{0.122(1-0.122)}{411} } } , 0.122+1.645 \times {\sqrt{\frac{0.122(1-0.122)}{411} } } ]

 = [0.095 , 0.148]

Therefore, 90% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating is [0.095 , 0.148].

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